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Part II  Multivariate Calculus

Chapter 11. Vectors and the Geometry of Space

Calculus

http://matrix.skku.ac.kr/Cal-Book1/Ch1/

http://matrix.skku.ac.kr/Cal-Book1/Ch2/

http://matrix.skku.ac.kr/Cal-Book1/Ch5/

Chapter 11. Vectors and the Geometry of Space

11.1 Three-Dimensional Coordinate Systems

문제풀이 by 김태현  http://youtu.be/_s_2T1VVob8

11.2 Vectors

문제풀이 by 오교혁  http://youtu.be/BFgh6irMqsc

11.3 The Dot Product

11.4 The Vector or Cross Product

11.5 Equations of Lines and Planes

문제풀이 by 구본우  http://youtu.be/lxuGE_Erthg

11.1 Three-Dimensional Coordinate Systems

In a two-dimensional plane any point can be represented as an ordered pair of real numbers, where is the -coordinate and is the -coordinate of the point. In three-dimensional space, a point is represented by three real numbers .

The coordinate axes labeled as the -axis, -axis and -axis are three directed lines that are perpendicular to each other and passing through a chosen fixed point (called the origin). This three-dimensional coordinate system (See Figure 1(a).) will enable us to represent a point in space. Normally, the -axis and -axis are taken as horizontal and the -axis as being vertical forming a Right-Handed Coordinate System, as in Figure 1(b). The three coordinate axes determine three coordinate planes. Thus the -plane contains the -axis and -axis; the -plane contains the -axis and -axis; the -plane contains the -axis and -axis. These three coordinate planes divide the space into eight parts, called octants. The positive axes determine the first octant, in the foreground. (See Figure 2.)

Figure 1 (a) Coordinate axes          (b) Right-hand rule

Figure 2 Octants

Let be any point in space. Then, the real numbers are the (directed) distances from to the -plane, the -plane, and the -plane respectively. The point is denoted by this ordered triple and , and are known as the coordinates of ; is the -coordinate, is the -coordinate, and is the -coordinate. Thus starting at the origin , the point is located by moving units along the -axis, then units parallel to the -axis, and then units parallel to the -axis as shown in Figure 3.

Figure 3

The three-dimensional rectangular coordinate system (Cartesian coordinate system) is the set of ordered triples given by the Cartesian product

.

Recall that an equation of and in two dimension represents a curve. Similarly, an equation of , and in three dimension represents a surface. In general, if is a constant, then represents a plane parallel to the -plane, is a plane parallel to the -plane, and is a plane parallel to the -plane in .

Distance Formula in Three Dimensions

The distance between the points and is

.

Example 1

Determine the equation of a sphere with radius and center .

Solution. A sphere is the set of all points whose distance from the center is . (See Figure 4.) So, any point on the sphere satisfies or .

Figure 4

Thus, the equation of the required sphere is

.          ■

If the center is the origin , then the equation of the sphere is

Example 2

Show that is the equation of a sphere, and find its center and radius.

Solution. We can rewrite the given equation in the form of an equation of a sphere if we complete the squares:

.

Comparing this equation with the standard form in the result of Example 1, we see that it is the equation of a sphere with center and radius .   ■

Example 3

What region in is represented by the following inequalities?

, .

Solution. The inequalities can be rewritten as , so they represent the points whose distance from the origin is at least 2 and at most 3. But we are also given that , so the points lie on or above the -plane. Thus, the given inequalities represent the region that lies between (or on) the spheres and and above (or on) the -plane. (See Figure 5.)

var('x, y, z')

p1=implicit_plot3d(x^2 + y^2 + z^2 == 4, (x,-5, 5), (y,-5, 5), (z,-5, 5), opacity=0.2, color="red")

p2=implicit_plot3d(x^2 + y^2 + z^2 == 9, (x,-5, 5), (y,-5, 5), (z,-5, 5), opacity=0.4)

p3=implicit_plot3d(z==0, (x,-5, 5), (y,-5, 5), (z,-5, 5), opacity=0.4, color="green")

show(p1+p2+p3)

Figure 5

Example 4

Draw the surface in . (See Figure 6.)

Solution.

var('x, y, z')

implicit_plot3d(x+y-5==0, (x, -5, 5), (y, -5, 5), (z, -5, 5),

color='blue', opacity=0.5)

Figure 6

11.1 EXERCISES (Three-Dimensional Coordinate Systems)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-11-1-Sol.html

1. Draw the surface in .

Solution.

var('x, y, z')

implicit_plot3d(2*x-3*y+z==1, (x, -3, 2), (y,-3,2), (z,-3,3))

2. Draw the surface in .

Solution.

var('x, y, z')

s1=implicit_plot3d(x^2-y^2==3, (x,-3,3), (y,-3,3), (z,-2,0.5), color='red', opacity=0.3)

s1

3. Find the lengths of the sides of the triangle with vertices , and . Is a right triangle? Is it an isosceles triangle?

Solution. Isosceles triangle

3

3

3*sqrt(2)

4. Find the distance from to each of the following.

(a) The -axis (b) The -axis

(c) The -axis (d) The -plane

(e) The -plane       (f) The -plane

Solution. Answer :  (3, 4, 5, sqrt(41), sqrt(34), 5)

5. Find the equation of the sphere with center and radius 3. What is the intersection of this sphere with the -plane?

Solution. The equation of the sphere :

, and the intersection of this sphere with the -plane can be obtained by substituting in the equation. Hence

.

6. Find an equation of the sphere that passes through the point and has center .

Solution. The distance between and  is the radius of the sphere.

Hence,

.

Thus, an equation of sphere is

.

7-8. Show that the equation represents a sphere, and find its center and radius.

7.

Solution.

8.

Solution. Completing squares in the equation gives :

, with the center

9. (a) Prove that the midpoint of the line segment from

to is , .

(b) Find the lengths of the medians of the triangle with vertices , and .

10-16. Determine the region of represented by the equation or inequality.

10. .

Solution. The equation represents a plane parallel to the -plane and 8 units in front of it.

11. .

12. .

Solution. The inequality represents all points on or between the horizontal planes (the -plane) and . So the answer is all points on or between the horizontal plane (the -plane) and .

13. , .

Solution.

var('x, y, z')

s1=implicit_plot3d(x^2+y^2==3, (x,-3,3), (y,-3,3), (z,-2,0.5), color='red', opacity=0.3)

s2=implicit_plot3d(z==-1, (x,-3,3), (y,-3,3), (z,-2,0.5), color='green', opacity=0.5)

s3=implicit_plot3d(x^2+y^2==3, (x,-3,3), (y,-3,3), (z,-0.99,-1.01), color='blue')

s1+s2+s3

14. .

Solution. The set of all points in whose distance from the -axis is . This is a cylinder of radius 3 and axis along -axis.

15. .

Solution. The inequality is equivalent to . So the region consists of those points whose distance from the point is greater than 1. This is the set of all points outside the sphere with radius 1 and center .

16. .

Solution.

var('x, y, z')

implicit_plot3d(x^2 + z^2 == 9 -2*z, (x, -5, 5), (y, -5, 5), (z, -5, 5), opacity=0.2, color="red")

17-18. Describe the given region by an inequality.

17. The half-space consisting of all points to the left of the -plane.

Solution. This describes all points with positive -coordinates, that is, .

18.The solid rectangular box in the first octant bounded by the planes , , and .

Solution.

var('x, y, z')

p1=implicit_plot3d(x == 1, (x, -5, 5), (y, -5, 5), (z, -5, 5), opacity=0.2, color="red")

p2=implicit_plot3d(y == 3, (x, -5, 5), (y, -5, 5), (z, -5, 5), opacity=0.2, color="blue")

p3=implicit_plot3d(z == 2, (x, -5, 5), (y, -5, 5), (z, -5, 5), opacity=0.2, color="green")

show(p1+p2+p3)

11.2 Vectors

Vectors are quantities that have both magnitude (or length) and direction.  Recall that a scalar is a real number (with only magnitude).

For example, velocity, acceleration, displacement of an object moving in space and a force are all vector quantities.

Figure 1

A vector is denoted graphically by an arrow or by a directed line segment. A vector is printed as a boldface letter or by putting an arrow above the letter.

The magnitude of the vector is the length of the arrow and the arrow points in the direction of the vector.

For example in Figure 1, the displacement of a particle along a line segment from point to point is represented by . This displacement vector v, has initial point (the tail) and terminal point (the tip).

Equality of vectors: Two vectors u and v are equal (or equivalent) and write uv if they have the same length and same direction (or parallel) but not necessarily the same initial point. The zero vector, denoted by 0: it has length and no direction.

Addition of vectors: The sum of two vectors u and v denoted by uv is the vector from the initial point of to the terminal point of as shown in the triangle law in Figure 2.

Figure 2 The Triangle Law              Figure 3 The Parallelogram Law

By placing v with the same initial point as u, we can construct a parallelogram with u and v as adjacent sides. Then, is the diagonal of the parallelogram starting at the same initial point. (This is called the Parallelogram Law: See Figure 3.) Note that uvvu. ( vector addition is commutative.) For instance, if several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces.

Scalar Multiplication: For any scalar and a vector v, the scalar multiple v is the vector whose length is times the length of v and whose direction is the same as v if and is opposite to v if . If , then (the zero vector). Scalar multiplication amounts to scaling the vectors, which can be elongation or contraction.

Figure 4

For example , known as negative of v, has the same length as v but points in the opposite direction. The difference of two vectors and is defined by .

Also if or .

Two non zero vectors are parallel if they are scalar multiples of one another.

Component Form of a Vector

Vectors can be treated algebraically by introducing a coordinate system. The position vector of a point is the vector with the initial point at the origin and terminal point at . The entries are called the components of and the vector was written as . The ordered triple refers to a vector while the ordered triple refers to a point in the space.

For any vector with the initial point and the terminal point , the vector representation of is by

.

For example, for the vector represented by the directed line segment with the initial point and terminal point , is

.

The magnitude (or length) of the vector v is denoted by the symbol or . (read as norm )

The length (or magnitude) of the position vector is    .

A unit vector is a vector whose length is 1. In general, if , then the unit vector that has the same direction as is

.

The vector is the normalization of the vector , has the same direction as and

In component form, vectors are added by just adding their corresponding components. Similarly, we can subtract vectors by subtracting the corresponding components and multiple of a vector by a scalar is done by multiplying each component by the scalar . Thus,

,

,

.

We denote by the set of all three dimensional vectors. More generally, denotes the set of all -dimensional vectors. An -dimensional vector is an ordered -tuple: having components . For -dimensional vectors, addition and scalar multiplication is defined in terms of its components just as for the cases and .

Algebra of Vectors

If and are vectors in and and are scalars, then

1.              2.

3.                  4.

5.          6.

7.              8.

9.                    10.

Equality of Vectors: Two vectors u and v are equal if and only if their corresponding components are equal.

Let , and be unit vectors in the directions of the positive -, -, and -axes. Thus unit vectors are write as follow:

Figure 5 Standard basis vectors in and

Any vector in can be expressed in terms of the standard basis vectors (or unit vectors along axes) , and . For , we can write

.

The above representation is unique. For example, .

For example, if and , then   .

For example, the unit vector in the direction of the vector is

Figure 6

. Note that the unit vector in the opposite direction of the vector is .

Application

Two forces and with magnitudes 10lb and 12lb act on an object at a point as shown in Figure 7. Find the resultant force acting at as well as its magnitude and direction. (Indicate the direction by finding the angle shown in Figure 7.)

Figure 7

We want to find . First we can find as

Hence,

Example 1

(a)Draw the vectors , and

(b) Show by means of a sketch, that there are scalars and such that

(c) Find the exact values of and .

Solution.

z=(0, 0); a=(-2, 3); b=(-2, -5); c=(7, 0)

s, t=var('s, t')

Ab = matrix(QQ, 2, 3, [a[0], b[0], c[0], a[1], b[1], c[1]])

T = Ab.echelon_form()

s=T[0, 2];t=T[1, 2]

A=arrow(z, a, color=(2, 1, 1))

A=A+text("a", (a[0]+0.3, a[1]+0.3))

A=A+arrow(z, (s*a[0], s*a[1]), color='blue')

A=A+text("sa", (s*a[0]+0.3, s*a[1]+0.3))

A=A+arrow(z, b, color=(3, 1, 0))

A=A+text("b", (b[0]+0.3, b[1]+0.3))

A=A+arrow(z, (t*b[0], t*b[1]), color='green')

A=A+arrow((t*b[0], t*b[1]), c, color='red')

A=A+text("tb", (t*b[0]+0.3, t*b[1]+0.3))

A=A+arrow((s*a[0], s*a[1]), c, color='red')

A=A+arrow(z, c, color='black')

A=A+text("c", (c[0]+0.3, c[1]+0.3))

A=A+point((a, b, c, (s*a[0], s*a[1]), (t*b[0], t*b[1])),  rgbcolor='brown', size=30)

show(A)

Figure 8          ■

Example 2

For given vectors , , and . Find scalars and such that .

Solution.

z=vector([0, 0,0]); a=vector([1, -1, 2]); b=vector([3, -1, 2]); c=vector([0,1,2]); d=vector([2,1,4])

Ad= matrix(QQ, [a, b, c, d]).transpose()

print T

print -1/4*a + 3/4*b + 3/2*c

p=plot(a,color='red')+plot(b,color='green')+plot(c)+plot(d,color='goldenrod')

p

var('r,s,t')

n=3

eq=[r*a[i]+s*b[i]+t*c[i]==d[i] for i in range(3)]

solve(eq,r,s,t)

[ 1  3  0  2]

[ 1  3  0  2]

[-1 -1  1  1]

[ 2  2  2  4]

[   1    0    0 -1/4]

[   0    1    0  3/4]

[   0    0    1  3/2]

(2, 1, 4) #  vector

[[r == (-1/4), s == (3/4), t == (3/2)]]

d = -1/4*a + 3/4*b + 3/2*c.                         ■

11.2 EXERCISES (Vectors)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-11-2-Sol.html

1-2. Determine .

1. (a) ,

(b) ,

Solution. (a)

(b)

v1=vector([5, -3])

v2=vector([10, 7])

v2- v1

2. ,

Solution. .

3-4. Find the sum of the given vectors.

3. ,

Solution. .

4. ,

Solution.

v1=vector([0, 1, -4])

v2=vector([0, 2, 0])

v1+v2

5-8. Compute and .

5. ,

Solution. ,

,

.

6. , .

Solution.

a=vector([5, -1, 3])

b=vector([-1, 3, -2])

a.norm()

a+b

2*a-3*b

(a-b).norm()

7.

Solution. ,

,

,

.

8.

Solution.

a=vector([3, -4, 0])

b=vector([1, -1, 1])

print a.norm()

print a+b

print 2*a-3*b

print (a-b).norm()

(4, -5, 1)

(3, -5, -3)

sqrt(14)

9. Determine a unit vector that has the same direction as .

Solution. The vector has length

, so the unit vector with the same direction is

.

10. Find a vector that has the same direction as but has length 6.

Solution.

v1=vector([2,-4,-2])

v2=6/v1.norm()*v1

print v2

v2.norm()

6

11. A clothesline is tied between two poles, 6m apart. The line is quite taut and has negligible sag. When a wet shirt with a mass of 0.8kg is hung at the middle of the line, the midpoint is pulled down 6cm. Find the tension in each half of the clothesline.

Solution. Let and represent the tension vectors in each side of the clothe line as shown in the figure. Then and have equal vertical components and opposite horizontal components, so

and .

By similar triangles, .

The force due to gravity acting on the shirt has magnitude , hence we have . The resultant of the tensile forces counterbalances , so

and .

Thus, the tensions are

.

12.The tension at each end of the chain has magnitude 50N. What is the weight of the chain?

13. (a) Draw the vectors , and .

(b) Show, by means of a sketch, that there are scalars and such that .

(c) Find the exact values of and .

Solution. (a)

a=vector([2, -3])

b=vector([-2, -1])

c=vector([6, -5])

p=plot(a, color='red')+plot(b, color='green')+plot(c)

t=text("a", (a[0]+0.3,a[1]+0.3))+text("b",(b[0]-0.3,b[1]+0.3))+text("c",(c[0]+0.3,c[1]+0.3))

show(p+t, figsize=4)

(b)

j=2

k=-1

z=(0, 0)

a=(2, -3)

b=(-2, -1)

c=(6, -5)

A=arrow(z, a, color=(2, 1, 1))+arrow(z, b, color=(3, 1, 0))+arrow(z, c, color='black')

A=A+arrow(z, (j*a[0], j*a[1]), color='blue')+arrow(z, (k*b[0], k*b[1]), color='green')

A=A+arrow((k*b[0], k*b[1]), c, color='red')+arrow((j*a[0], j*a[1]), c, color='red')

A=A+text("a", (a[0]+0.3, a[1]+0.3))+text("ja", (j*a[0]+0.3, j*a[1]+0.3))+text("b", (b[0]+0.3, b[1]+0.3))

A=A+text("kb", (k*b[0]+0.3, k*b[1]+0.3))+text("c", (c[0]+0.3, c[1]+0.3))

A=A+point((a, b, c, (j*a[0], j*a[1]), (k*b[0], k*b[1])), rgbcolor='brown', size=30)

show(A)

var('s,t')

a=vector([2,-3])

b=vector([-2,-1])

c=vector([6,-5])

n=3

eq=[s*a[i]+t*b[i]==c[i] for i in range(2)]

solve(eq,s,t)

14. Let , , and in .

(i) Plot the vectors and .

(ii) Find scalars , and such that .

Solution.

a=vector([2,-3,1])

b=vector([-2,-2,1])

c=vector([6,-5,2])

d=vector([2,1,2])

p=plot(a,color='red')+plot(b,color='green')+plot(c)+plot(d,color='goldenrod')

p

var('r,s,t')

n=3

eq=[r*a[i]+s*b[i]+t*c[i]==d[i] for i in range(3)]

solve(eq,r,s,t)

Answer :  [[r == (-28/3), s == (8/3), t == (13/3)]]

15. If and , describe the set of all points such that .

Solution.

Therefore the surface of a sphere with a center and a radius .

11.3 The Dot Product

The product of two vectors which is known as a scalar product or dot product or even as inner product, is defined as follows:

The dot product of and is the real number (scalar) given by .

The dot product of and is obtained by multiplying corresponding components and then adding the individual products.

For example,

,

.

Properties of the Dot Product

If , and are vectors in and is a scalar, then

(ⅰ)

(ⅱ)

(ⅲ)

(ⅳ)

(ⅴ)

Proof.  Let , and . We shall prove few of them and remaining as exercise.

(ⅰ)

(ⅲ)

■

Geometric interpretation: The dot product enables us to find the angle between two nonzero vectors. Let be angle between and where in Figure 1.

Figure 1

Then the dot product of two nonzero vectors and is . Thus, the formula of the angle between two nonzero vectors and is given by

.

Figure 2

The dot product of two vectors is itself a scalar. Two special cases immediately arise:

(i) and are perpendicular if and only if .

(ii) and   are parallel if and only if .

The zero vector 0 is perpendicular to all vectors.

Specially, for the unit vectors in ,

.

For example, consider the vectors and having lengths 5 and 8 with an angle between and .

As a result, we have the dot product of and is .

Example 1

Find .

Solution.

a=vector(QQ, [2, -3]);b=vector(QQ, [1, -7])

ab=a.dot_product(b)

Example 2

Find the angle between the vectors and .

Solution. Since,

and .

We have

That is, the angle between and is

def anglebetween(a,b)

return arccos(a.dot_product(b)/(a.norm()*b.norm()))

a=vector([2, 4, -4])

b=vector([2, 2, 0])

anglebetween(a, b)

Example 3

Show that is perpendicular to .

Solution. Since,

these vectors are perpendicular.      ■

Since if and if , it follows that is positive for and negative for . The dot product measures the extent to which and point in the same direction.

If , then and point in the same general direction, if , then they are perpendicular, and if , then they point in generally opposite directions. (See Figure 3.) When and point in exactly the same direction, we have , so and .

If and point in exactly opposite directions, then and so and

.

Figure 3

Proposition: Let  , and be vectors. If and   are both perpendicular to , then every linear combination is perpendicular to .

Proof.  Suppose that and are both perpendicular to . Then

and   .

It follows that and therefore is perpendicular to .         ■

Direction Angles and Direction Cosines

The angles , and which a nonzero vector a makes with the positive -axis, -axis and -axis are known as the direction angles of a. These angles lie in the interval .

Figure 4

The direction cosines of a vector a are the cosines of these direction angles and that are , and , respectively.

Thus we have

, , .

Consequently,

.

Also,

.

Therefore, dividing by both sides,

.

Thus, the direction cosines of are the components of the unit vector in the direction of .

Example 4

Find the direction angles of the vector

Solution. Since , gives

,   ,   .

Hence

.   ■

Projections and Components

Consider two vectors and with the same initial point and with as the angle between them, represented by and . Let be the foot of the perpendicular from to the line containing . Then is called the vector projection of onto and is denoted by .

Figure 5  Vector projections

The component of b along a (or the scalar projection of onto ) is the signed magnitude of the vector projection, which is the number . We denote this by . Observe that it is negative if . Since

the dot product of and is the product of the magnitude of and the scalar projection of b on to . Also

the component of along is obtained by the dot product of with the unit vector in the direction of . Thus, as a summary, the followings are true:

Scalar projection of onto : ,

Vector projection of onto :

The vector projection is the scalar projection times the unit vector in the direction of a.

(a)                         (b)

Figure 6

For each vector

,

This agrees with our previous use of the term “component” and gives the identity

.

The vector is orthogonal to the projection vector .

Hence the equation

.

Note that is orthogonal to the vector .

Example 5

Find the scalar projection and vector projection of onto .

Solution. Since , the scalar projection of onto is

.

The vector projection is this scalar projection times the unit vector in the direction of . That is,

.

a=vector([1,5,-3])

b=vector([2,3,-1])

(a.dot_product(b))/(a.norm()^2)*a

Answer :  (4/7, 20/7, -12/7)     ■

Some Applications

When a force is directed along the line of motion of the object, then the work done by in moving the object through a distance is . Suppose that the constant force , pointing in some other direction (See Figure 7.), moves the object from to , then the displacement vector is . Then, the work done by this force is defined to be the product of the component of the force along and the distance moved. If is the angle between and , then

.

If causes a displacement of a body, then the work done is

.

Figure 7

Example 6

A crate is hauled 10m up a ramp under a constant force of 160N applied at an angle of to the ramp. Find the work done.

Figure 8

If and are the force and displacement vectors, as pictured in Figure 8, then the work done is

.   ■

A constant force with vector representation moves an object along a straight line from a point to anthor point. The work done when we find in this book the distance is measured in meters and the magnitude of the force is measured in newtons.

Example 7

Find a work done when a force and the displacement vector is

Solution.

def proj_ab(a,b):

return (a.dot_product(b))/(a.norm()^2)*a

a=vector([1,5,-3])

b=vector([2,3,-1])

proj_ab(a,b)

a=vector(QQ, [2, -3]);b=vector(QQ, [1, -7])

ab=a.dot_product(b)

show(ab)

11.3 EXERCISES (The Dot Product)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-11-3-Sol.html

1. Determine the dot product of two vectors if their lengths are 8 and and the angle between them is .

Solution. Let the vectors be and Then,

by definition of the dot product.

2-6. Find the dot product of and

2. , .

Solution.

a=vector([5, -3]);

b=vector([4, 6]);

a.dot_product(b);

3. .

Solution.

.

4. , .

Solution.

5. .

Solution.

6. and the angle between and is .

Solution. .

7-9. Compute the angle between the vectors.

7. , .

Solution. and

.

From the definition of the dot product, we have

.

Hence, the angle between and is

.

That is, and are orthogonal.

8. (a) ,

(b) ,

Solution. (a)

def anglebetween(a,b):

return arccos(a.dot_product(b)

/(a.norm()*b.norm()))

a=vector([6,3])

b=vector([4,2])

anglebetween(a,b)

(b)

a=vector([6, -3, 4]);

b=vector([2, 0, -3]);

anglebetween(a,b)

9. , .

http://matrix.skku.ac.kr/cal-lab/11-3-9.html

Solution. ,

, and .

From the definition of the dot product, we have

and

.

a=vector([2, -1, 3])

b=vector([3, 1, -5])

print "The angle between a and b is", anglebetween(a,b).n(), "radians"

print "The angle between a and b is", anglebetween(a,b).n()*180/pi.n(), "degrees"

The angle between a and b is 2.03952668853669 radians

The angle between a and b is 116.856271457445 degrees

10. Verify whether the given vectors are orthogonal, parallel, or neither.

(a) ,

(b) ,

(c) ,

Solution. (a) Since , and are orthogonal.

(b)

def anglebetween(a,b):

return

arccos(a.dot_product(b)/(a.norm()*b.norm()))

var('t')

a=vector([6, 3]);

b=vector([4, 2]);

solve(a.dot_product(b)/(a.norm()*b.norm())==cos(t), t)

anglebetween(a,b)

(c) Parallel.

11. Determine such that the vectors and are orthogonal.

Solution.

var('t')

a=vector([2, -6, t])

b=vector([t, t, t^2])

x=a.dot_product(b)

solve([x==0], t)

Answer :  [t == -2, t == 2, t == 0]

12. Find a unit vector that is orthogonal to both and .

Solution. .

13-14. Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.)

13.

Solution. Since , the direction cosines of the vector are

.

Hence, .

14. .

Solution. direction cosine: ,

direction angle: , , .

15.Prove that the vector known as orthogonal projection of , is orthogonal to .

Solution.

This proves the result.

16-19. Find the scalar and vector projections of onto and orthogonal projection of , and .

16. , .

Solution. scalar projection: ,

vector projection: ,

orthogonal projection: .

17. , .

Solution.

a=vector(QQ, [2, -1, -2])

b=vector(QQ, [4, 3, 3])

ab=a.dot_product(b)

an=a.norm()

scal_proj=ab/an

scal_proj

vec_proj=scal_proj/an*a

vec_proj

orth_proj=b-vec_proj

orth_proj

vec_proj.dot_product(orth_proj)

18. ,

Solution. scalar projection: ,

vector projection: ,

orthogonal projection: .

19. ,

Solution. ,

so

and .

And .

20.Prove that the distance from a point to the line is

.

Find the distance from the point to the line .

21. Prove the Cauchy-Schwarz Inequality:  .

Solution. Since ,

.

22. Prove the Triangle Inequality:  .

Solution.   Note that it is enough to prove .

Consider the  L․H․S ,

Hence we have

23. Prove that

.

Solution.

and .

.

24.Show that vectors a and b are orthogonal if and only if .

25.Show that if and only if is orthogonal to .

26.Give geometric interpretation of the above two problems.

11.4 Cross Product

The cross product is only defined for vectors in and results in another vector in .

Component Form of the Cross Product

We define the cross product of two vectors and in the terms of the components of the vector.

THEOREM 1 Cross Product of Two Vectors

The cross product of two vectors and is the vector

.

The cross product of the vectors and can also be obtained as follows

or in the determinant form

.

For example, we consider two vectors and . Then

.

Example 1

Find a vector perpendicular to the plane that passes through the points , and .

Solution. The vector is perpendicular to both and and is therefore perpendicular to the plane through and . We know that

We compute the cross product of these vectors:

Hence, the vector is perpendicular to the given plane. Any nonzero scalar multiple of this vector, such as is also perpendicular to the Plane Equation.

Solution. 1. Note that and . (use determinant form for simplicity) Thus is a vector which is orthogonal to both and .

2. Suppose be a vector which is orthogonal to vectors and . Then components of can be obtained by solving two equations and .

3. , that is the vector product is not commutative. (In fact, )

Right-Hand Rule

An alternative characterization of the cross product uses the right-hand rule. As seen in Figure 1, if the fingers of the right hand point along the vector and then curl toward the vector , the thumb will give the direction of . In Figure 1, the right-hand rule shows the direction of .

Figure 1

THEOREM 2

Let and be two vectors in . Then the cross product is a vector with magnitude

where is the angle between and . The direction of is the direction given by the right-hand rule. (See Figure 1.) When , the direction of is undefined.

Proof.  If and , then .

It is easy to see that

Taking square root both sides and observing that

for ,

we get the result.

Area

Let two nonzero vectors and be two sides of a parallelogram, then the area of the parallelogram is

.

Likewise, we see that the area of a triangle with sides and is

.

Thus, the length of the cross product is equal to the area of the parallelogram (See Figure 2.) determined by and .

Figure 2

Example 2

Find the area of the triangle with vertices , and .

Solution. In Example 1, we computed . The area of the parallelogram with adjacent sides and is the length of this cross product:

The area of the triangle is half the area of this parallelogram. Hence the required area of its triangle is .   ■

The following theorem is a consequence of the definition of the cross product.

THEOREM 3  Parallel Vector

Two nonzero vectors and are parallel ( or ) if and only if .

Using , then it follows that for the standard basis vectors , , and , we have

,         ,       ,

,      ,              .

Observe that .

In general, . Hence, the associative law for multiplication usually does not hold. For example whereas .

For vector products, the following usual laws of algebra hold.

THEOREM 4  Properties of the Cross Product

If , and are vectors and is a scalar, then

(ⅰ)

(ⅱ)

(ⅲ)

(iv) or

(v)

(vi)

(vii)

Readers are encouraged to verify the above properties by using components of vectors.

Triple Scalar Product or Box Product

If , and , then

.

The product is called the triple scalar product of , and .

Note that follows from the properties of determinants.

http://matrix.skku.ac.kr/2012-LAwithSage/interact/1.html

Volume of a Parallelepiped

Consider a parallelepiped with coterminous edges determined by the vectors , , and . (See Figure 3.)

Figure 3

The area of the base parallelogram is . If is the angle between a and , then the height of the parallelepiped is . Therefore, the volume of the parallelepiped determined by the coterminous edge vectors , and is

the magnitude of the scalar triple product.

Example 3

Find the volume of the box (parallelepiped) determined by , and .

Solution. The volume of the parallelepiped is given by

.

Therefore, the volume is units cubed.

a=vector([3, 1, 1])

b=vector([1, 4, 1])

c=vector([1, 1, 5])

a.dot_product(b.cross_product(c))

THEOREM 5

The three vectors , , are coplanar (lie in the same plane) if and only if the volume of the parallelepiped is zero, and consequently .

Example 4

Use the scalar triple product to show that the vectors , and are coplanar.

Solution. We use to compute their scalar triple product:

.

Therefore, by Theorem 5 the volume of the parallelepiped determined by , and is 0. This means that , and are coplanar.   ■

Torque

Suppose a force is acting on a rigid body at a point given by a position vector . Let be the angle between the position and force vectors. The torque vector with reference to the origin is .

It measures the tendency of the body to rotate about the origin. Its direction indicates the axis of rotation. Its magnitude is equal to the area of the parallelogram determined by and .

For example, if a bolt is tightened by applying a force to a wrench (See Figure 4.), it produces, a turning effect. Note that the only component of that can cause a rotation is the one perpendicular to , that is, .

Figure 4

Example 5

A bicycle pedal is pushed by a foot with a 40N force as shown in Figure 5. The shaft of the pedal is 20cm long. Find the magnitude of the torque about .

Figure

Solution.

.   ■

Example 6

Find the cross product and verify that it is orthogonal to both and :

and

Solution.

a=vector(QQ, [1, 1, -2]);b=vector(QQ, [1, 0, -1])

c=a.cross_product(b)

show(c)

show(e.dot_product(a))

show(e.dot_product(b))

Answer :  (-1, -1, -1)  # a.cross_product(b)

0         # (e.dot_product(a))

0            # (e.dot_product(b))       ■

11.4 EXERCISES (The Cross Product)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-11-4-Sol.html

1-5. Find the cross product and verify that it is orthogonal to both and .

1. , .

Solution.

a=vector(QQ, [1, -1, 1])

b=vector(QQ, [2, 0, 3])

c=a.cross_product(b)

show(c)

show(e.dot_product(a))

show(e.dot_product(b))

0

0

2. , .

Solution.

Now, and

So, is orthogonal to both and .

3. , .

Solution. .

4. .

Solution. .

5. .

Solution. .

6. If and , find and .

Solution. , .

7.If , and ,

show that .

Solution.

(i)

(ii)

Hence, .

8. Find two unit vectors orthogonal to both and .

Solution.

a=vector(QQ, [1, -1, 2])

b=vector(QQ, [3, 0, 1])

c=a.cross_product(b)

cn=c.norm()

e=c/cn

show(e)

show(-e)

(1/35*sqrt(35), −1/7*sqrt(35), −3/35*sqrt(35))

. Thus, two unit vectors orthogonal to both are

,

that is, and  .

9. Find two unit vectors orthogonal to both    and .

Solution. .

Thus, two unit vectors orthogonal to both are , that is, and .

10. Find the area of the parallelogram with vertices , , and

.

Solution. We may think of these points in -plane in the space.

A=vector([0, 1, 0])

B=vector([2, 1, 0])

C=vector([1, 4, 0])

D=vector([1, -2, 0])

AB=B-A; AC=C-A

Area= abs(AB.cross_product(AC))

show(Area)

11.Find the area of the parallelogram with vertices , , and .

Solution. The parallelogram is determined by the vectors and , so the area of parallelogram is

Then .

12-13. Find a vector perpendicular to the plane through the points , and .

12. , ,

Solution.

P=vector([1, 0, 0]);

Q=vector([4, 1, -1]);

R=vector([2, -1, -2]);

PQ=Q-P;

PR=R-P;

PQ.cross_product(PR)

13. , ,

Solution. and , so a vector orthogonal to the plane through and is

.

That is, is orthogonal to the plane through and .

14-15. Find the area of triangle .

14. , ,

Solution.

P=vector([1, 0, 0]);

Q=vector([4, 1, -1]);

R=vector([2, -1, -2]);

PQ=Q-P;

PR=R-P;

CP=PQ.cross_product(PR);

1/2*CP.norm()

15. , , .

Solution. .

16-17. Find the volume of the parallelepiped with adjacent edges , , and .

16. , , ,

Solution.

P=vector([2, 0, 1]);

Q=vector([4, 2, 0]);

R=vector([3, 1, -1]);

S=vector([1, 1, 0]);

PQ=Q-P;

PR=R-P;

PS=S-P;

CP=PQ.cross_product(PR);

PS.inner_product(CP).abs()

17.,,,.

Solution. and .

.

So, the volume of the parallelepiped is cubic units.

8.Show that the vectors , and are not coplanar.

Solution. Not coplanar.

a=vector(QQ, [1, 1, 0])

b=vector(QQ, [2, -1, 4])

c=vector(QQ, [2, 1, 4])

M=matrix(QQ, [a, b, c]);M

print M

print M.det()

Answer : [ 1  1  0]

[ 2 -1  4]

[ 2  1  4]

-8

19.Determine whether the points , , and lie in the same plane.

Solution. and .

.

Thus, the volume of the parallelepiped determined by and is . This says that these vectors lie in the same plane. Therefore, their initial and terminal points and also lie in the same plane.

20. A wrench 40cm long lies along the positive -axis and grips a bolt at the origin. A force is applied in the direction at the end of the wrench. Find the magnitude of the force needed to supply of torque to the bolt.

21.Suppose that . Prove or disprove the following statements.

(a) If , then

(b) If , then

(c) If and , then

Solution. (a) False.

If , then , hence is perpendicular to . This can happen if .

For example, let and , then .

(b) False.

If , then , which implies that is parallel to , which of  course can happen if .

(c) True.

Since , is perpendicular to , by part (a). From part (b), is parallel to .  Since , and is both parallel and perpendicular to , we have . Hence .

22. Show that .

23. If and , then find .

11.5 Equations of Lines and Planes

In this section, we will derive vector equations of lines and planes in and , and we will deal with shortest distance problems related to these equations.

Lines

In the plane , the equation of the line can be uniquely determined when a slope and a specified point on the line are given. In general it can be generally written as follows:

where , and are real numbers and and are not both zero.

Let's find the equation of a line in . If a line passes through the point and is parallel to , then the vector is parallel to   (See Figure 1.), where is any point on the line.

Figure 1

That is, the line is a set of all points that satisfies the following equation :

Thus, if is any point on the line through that is parallel to , then the vector is parallel to , so for some . This line can be represented as the equation .

We call this a vector equation of the line through that is parallel to .

A vector equation of a line can be split into a set of scalar equations by equating corresponding components; these are called parametric equations of the line. Thus we have vectors

,

which implies

From , symmetric equations of the line can be written defined as the following:

where , and are nonzero constants.

If and are distinct points with position vectors and in or , then the line determined by these points is parallel to the vector , so it follows from that the line can be expressed in vector form as

, .

Equation is called the two-point vector equation of the line through and .

Example 1

(a) Find a vector, parametric and symmetric equations of the line that pass through the point and is parallel to the vector .

(b) Find two other points on the line.

Solution. (a)Here and  .

The vector equation is

.

Parametric equations are , , .

The symmetric equation is

.

(b)Choosing the parameter value gives , , and , so is a point on the line. Similarly, gives the point .   ■

Example 2

Find a vector, parametric and symmetric equations for the line that pass through the points and .

Solution. Two points and with position vectors and forms a vector

and the vector equation can be written  as

, .

Thus, the parametric equations are , , and symmetric equations of the line are  .

Point-Normal Equation of Planes

A plane in can be uniquely obtained by specifying a point in the plane and a nonzero vector that is perpendicular to the plane. (See Figure 2)

Figure 2

The vector is called the normal vector to the plane. If is any point in this plane, then the is orthogonal to (See Figure 2). By the property of the dot (inner) product

From ,

or

where , and are not all zero.

We call a point-normal equation of the plane through with normal

.

For convenience, we simplify the left terms of as follows

where , and are not all zero and .

We call the general equation of the plane.

Example 3

Find a point-normal equation and a general equation of the plane that passes through with normal .

Solution. From , a point-normal equation of the plane is

.

Multiplying out and taking the constants to the right side yields the general equation

.      ■

Vector and Parametric Equations of Planes

A plane can be uniquely obtained by passing through a point in and two nonzero vectors and that are parallel to and are not scalar multiples of one another. (See Figure 3.) That is, if is any point in the plane and and are positioned with their initial points at , then is expressed as a linear combination of and ;

or

where and and and , called parameters, are in . We call a vector equation of the plane through that is parallel to and .

Let be any point in the plane through that is parallel to the vectors and . Then can be expressed in the component form as

or

We call parametric equations of the plane.

Figure 3

Example 4

Find vector and parametric equations of the plane that passes through three points: , , .

Solution. Let , , and . Then we have two vectors that parallel to the plane as

, .

From , a vector equation of the plane is

.

Also we have parametric equations as

.      ■

Perpendicular and Parallel Lines

Two lines and with direction vectors and , respectively, are

(i) intersect,

(ii) parallel if for some nonzero scalar ,

(iii) skew if neither (i) nor (ii).

Example 5

Show that the lines

: , , ,

: , ,

are perpendicular.

Solution. By taking the value for the parameters and , we get two vectors

and

are parallel to and , respectively. We can verify that two lines are perpendicular to each other by

.     ■

Example 6

Show that the lines

,

are parallel.

Solution. The coefficient of the parameters and , we see that

and

are the direction vectors for and , respectively. Since , the two lines are parallel.     ■

Example 7

Show that the lines and with parametric equations

: ,

:

are skew lines; that is, they do not intersect and are not parallel (and therefore do not lie in the same plane).

Figure 4

Solution. The lines are not parallel because the corresponding direction vectors and are not parallel. (Their components are not proportional.) If and have a point of intersection, there would be values of and such that

,

,

.

If we solve the first two equations, we get and , and these values do not satisfy the third equation. Therefore, no values of and satisfy all three equations. Thus and do not intersect. Hence, and are skew lines.

var ('t, s')

L1=parametric_plot3d([t,2*t+1,3*t +2], (t,-3.5,3.5), color='blue', thickness=5)

L2=parametric_plot3d([3-4*s,2-3*s,1+2*s], (s,-3.5,3.5), color='red', thickness=3)

show(L1+L2)

The Distance from a Point to a Line in Space

To find the distance from a point to a line that passes through a point parallel to a vector , we can find the relationship of the distance and the length as follows: (See Figure 5.)

Figure 5

where is the angle between and .

From Theorem 2 in Section 11.4,

which implies .

Then the distance from a point to a line is .

Example 8

Find the distance from the point to the line

: .

Solution. We see from the equations for that it passes through and is parallel to . We obtain

and

.

.        ■

Projection of Vectors

To find the distance from a point to a plane in , we introduce the concept of the orthogonal projection of vectors.

(a)                        (b)

Figure 6

Figure 7

Let be the foot of the perpendicular from to and be vectors in with . Then is called an orthogonal projection of onto . The vector can be expressed as . Here is called a vector component of that is perpendicular to . In Figure 7,  a vector is sum of and ;     .

The following theorem give expression for the orthogonal projection .

THEOREM 1

If and are in , then we have the following:

(a)            (b)

Proof. (a) Since is parallel to , we have .

Since and because is perpendicular to (See Figure 6), we can get the value as  that satisfies

.

Therefore .

(b) .

Example 9

Find the projection of onto and a vector component of perpendicular to for and .

Solution. Since and , the projection of onto is

and a vector component of perpendicular to is

.    ■

The Distance from a Point to a Plane in Space

We find a way to determine the distance from a point to the plane . Note that is the normal to the plane.

Figure 8

Let be any point in the given plane and let . Then, .

The distance from to the plane is equal to the absolute value of the scalar projection of onto the normal vector . (See Figure 8.) Thus,

.

Since lies in the plane, its coordinates satisfy the equation of the plane, so we have . Thus,

.

Example 10

Find the distance from a point to the plane .

Solution. Since a normal vector is ,

.   ■

Similarly, in the distance from to a line is

.

Example 11

Find the distance from a point to the plane .

Solution. Since a normal vector is ,

.

Now we have some examples for properties of lines and planes in .

Example 12

Find an equation of the plane through the point with the normal vector .

Solution. Putting , , , , and , from , we see that an equation of the plane is

or .    ■

Example 13

Find an equation of the plane that passes through the points , and .

Solution. The vectors and corresponding to and are

.

Since both and lie in the plane, their cross product is orthogonal to the plane and can be taken as the normal vector. Thus,

With the point and the normal vector , the equation of the plane is

or .    ■

Example 14

Find the point at which the line , , intersects the plane .

Solution. We substitute the expressions for , and from the parametric equations into the equation of the plane:

This simplifies to , so . Therefore, the point of intersection occurs when the parameter value is .

Thus, , , .

Hence the point of intersection is .   ■

Example 15

(a) Find the angle between the planes and .

(b) Find symmetric equations for the line of intersection of these two planes.

Solution. (a)The normal vectors of these planes are and , respectively.

Let be the angle between the two planes, which is called a dihedral angle. This dihedral angle is equal to the angle between two normal vectors , . By the property of the dot product,

,

and . Therefore two planes are orthogonal to each other.

(b) We first need to find a point on . For instance, we can find the point where the line intersects the -plane by setting in the equations of both planes. This gives the equations and , whose solution is , . Hence, the point lies on .

Now, we observe that, since lies in both planes, it is perpendicular to both of the normal vectors. Thus, a vector v parallel to is given by the cross product

.

Hence, the symmetric equations of can be written as

.

The line of intersection can also be obtained by solving the equations of the planes for two of the variables in terms of the third, which can be taken as the parameter.

For instance, the line was given as the line of intersection of the planes and . The symmetric equations that we found for could be written as

Figure 9 Shows how the line in Example 17 can also be regarded as the line of intersection of planes derived from its symmetric equations (Sage).

and

which is again a pair of linear equations. They exhibit as the line of intersection of the planes and . (See Figure 9.)

var ('x, y, z')

P1=implicit_plot3d(x + y + z == 1,(x,-7,7),(y,-7,7),(z,-7,7),color='blue',opacity=0.3)

P2=implicit_plot3d(x - 2*y + 3*z == 1,(x,-7,7),(y,-7,7),(z,-7,7),color='red',opacity=0.3)

show(P1+P2)

Perpendicular and Parallel Planes

Two planes and with normal vectors and , respectively, are

(i) perpendicular : if , and

(ii) parallel if for some nonzero scalar .

Example 16

Show that the planes are parallel and find the distance between the parallel planes and .

Solution. First, we note that the planes are parallel because their normal vectors and are parallel. To find the distance between the planes, we choose any point on one plane and calculate its distance to the other plane. In particular, if we put in the equation of the first plane, we get , so is a point in this plane. Then, the distance between and the plane is

.

Hence, the distance between the planes is .  ■

Example 17

In Example 7, we showed that the lines

: ,

: .

are skew. Find the distance between them.

Solution. Since the two lines and are skew, they can be viewed as lying on two parallel planes and . The distance between and is the same as the distance between and which can be computed as in Example 16. The common normal vector to both planes must be orthogonal to both (the direction of ) and (the direction of ). Hence, a normal vector is

.

If we put in the equations of we get the point on , so, an equation for is

or .

If we now set in the equations for , we get the point on . So the distance between and is the same as the distance from to . Then, this distance is

.     ■

Example 18

Find the distance from the point to the plane .

Solution. Use to get the distance.

a=vector(QQ, [2, -4, 3]);

d= -2

p=vector(QQ, [-3, 1, 5])

dis=abs(a.dot_product(p)+d)/a.norm()

show(dis)

Example 19

Plot the two planes and . Find the line of intersection of the two planes and plot it along with the two planes. (See Figure 10.)

http://matrix.skku.ac.kr/cal-lab/cal-11-5-19.html

Figure 10

Solution.

var ('x,y,z')

P1=implicit_plot3d(x + y + z == 1,(x,-2,3),(y,-2,3),(z,-2,3),color='orange',opacity=0.3)

P2=implicit_plot3d(2*x - y + z == 2,(x,-2,3),(y,-2,3),(z,-2,3),color='green',opacity=0.3)

show(P1+P2)

We may also get the parametric equation of the line of intersection of the two planes in Sage.

var('x, y, z')

solve([x+y+z==1, 2*x-y+z==2], [x, y, z])

11.5 EXERCISES (Equations of Lines and Planes)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-11-5-Sol.html

1-7. Find a vector equation, parametric equations and symmetric equations for the line.

1. Through the point and parallel to the vector .

Solution. For this line, we have and . Hence a vector equation is and parametric equations are . The symmetric equations are .

2. Through the point and parallel to the vector .

Solution.

var('t')

r0=vector([3, 5, 1])

pv=vector([4, 1, -1])

r0+t*pv

Answer :  (4*t + 3, t + 5, -t + 1)

parametric equation: , , ,

symmetric equation: .

3. Through the origin and parallel to the line , , .

Solution. This line has the same direction as the vector, .

Here , so a vector equation is and parametric equations are . The symmetric equations are .

4. Through the point and perpendicular to the plane .

Solution. This line is passing through and along the normal vector to the given plane.

(ℝ)

⇒ vector equation: ,

parametric equation: , , ,

symmetric equation: .

5. Through the origin and the point .

Solution. For this line, we have and . Hence a vector equation is and parametric equations are .

The symmetric equations are .

6.Through the points and .

Solution. parametric equation: , , ,

symmetric equation: .

var('t')

A=vector([3, 5, -3])

B=vector([-1, 0, 5])

v=B-A

A+t*v

Answer :  (-4*t + 3, -5*t + 5, 8*t - 3)

7. Through and perpendicular to both and .

Solution. A line perpendicular to the given two vectors has the same direction as a cross product of the two vectors. That is,

.

Here, , so a vector equation is and parametric equations are . The symmetric equations are .

8. Is the line through and parallel to the line through and ?

Solution. The lines are not parallel because the corresponding vectors , are not parallel.

var('t, s')

A=vector([3, 4, 5])

B=vector([-2, 0, 1])

v=B-A

print A+t*v

C=vector([2, 1, 4])

D=vector([-3, -3, -3])

w=D-C

print C+s*w

A=vector([3,4,5])

B=vector([-2,0,1])

C=vector([2,1,4])

D=vector([-3,-3,-3])

L1=line3d([A,B])

L2=line3d([C,D],color='red')

show(L1+L2)

Answer :   (-5*t + 3, -4*t + 4, -4*t + 5)

(-5*s + 2, -4*s + 1, -7*s + 4)

9.  Is the line through and perpendicular to the line through and ?

Solution. Direction vectors of the lines are and .

Since , the vectors and the lines are not perpendicular.

10-13. Determine whether the lines and are parallel, skew, or intersecting. If they intersect, find the point of intersection.

10. : , , : , , .

Solution.   It is apparent that the lines are skew in the following figure.

var('t, s');

L1=parametric_plot3d((25, 1+3*t, -4*t), (t, -50, 50))

L2=parametric_plot3d((3+s, 4-2*s, s), (s, -50, 50), color="red")

show(L1+L2)

11.: , , , ℝ.

: , , , ℝ.

Solution. Since the direction vectors are and , we have . Hence the lines are parallel.

12. : : .

Solution. The lines are not parallel because the corresponding vectors , are not parallel. If and have a point of intersection, there would be values of and such that

, ,   .

and are skew.

13. : : .

Solution. From the figure it is clear that the lines are intersecting.

var('x');

diff(x^(15/14)+5*e^x,x)

var('s, t')

A=vector([3,1,4])

B=vector([-2,2,1])

C=vector([4,3,7])

D=vector([3,0,2])

Lt=A+t*B

Ls=C+s*D

print "Clearly the two lines are intersecting"

parametric_plot3d(Lt,(t,-3,3))+parametric_plot3d(Ls,(s,-3,3),color='red')

sol=solve([Lt[0]==Ls[0], Lt[1]==Lt[1],Lt[2]==Ls[2]], s, t, solution_dict=True)

show(sol)

14-15. Find an equation of the plane.

14. Through the point and perpendicular to the vector .

Solution.   .

15.Through the point and with normal vector

Solution. is a normal vector to the plane and is a point of the plane. Then or to be the equation of the plane.

16. Which of the following four planes are parallel?

,   ,

.

Solution. and are parallel.

17. Which of the following four lines are parallel?

,    ,

,             .

Solution. and are parallel.

18-19. Find an equation of the plane through the given point with the normal vector which is the direction of the line with the given parametric equations.

18. , , .

Solution. is a normal vector to the plane and is a point of the plane. Then  or to be the equation of the plane.

19. , , .

Solution. is a normal vector to the plane and is a point of the plane. Then or is the equation of the plane.

20-21. Find the distance from the point to the given plane.

20. .

Solution.   The normal vector to the plane is

n=vector(QQ, [2, -1, 3])

d= -4

p=vector(QQ, [3, 1, 5])

dis=abs(n.dot_product(p)+d)/a.norm()

dis

21. .

Solution. The distance  .

22-23. Find the distance between the given parallel planes.

22. .

Solution. : , : . Note that is a point of the first plane. Since the planes are parallel, the distance between the two planes is the distance from to the  second plane. Then, the distance between and the  plane is

.

23. .

Solution. Put in the equation of the first plane to get the point on the plane. Since the planes are parallel, the distance between the two planes is the distance from to the  second plane. Hence

.

24.Find the distance between the two skew lines

and .

25. Prove that the distance between the parallel planes and is .

26. (Line of intersection of two planes)  Plot the two planes and . Find the line of intersection of two planes and hence plot this.

http://matrix.skku.ac.kr/cal-lab/cal-11-5-26.html

Solution.

var('x,y,z,t')

solve([x + y + z == 1,2*x - y + z == 2],x,y,z)

Answer :  x == -2/3*r1 + 1, y == -1/3*r1, z == r1

Clearly the line of intersection of the is .

P1=implicit_plot3d(x + y + z == 1, (x,-3,3), (y,-3,3), (z,-3,3), color='blue', opacity=0.3)

P2=implicit_plot3d(2*x - y + z == 2, (x,-3,3), (y,-3,3), (z,-3,3), color='red', opacity=0.3)

L=parametric_plot3d([1-2/3*t,-1/3*t,t], (t,-3.5,3.5), color='green',thickness=3)

show(P1+P2+L)

27. (Line of intersection of two planes)  If the two lines have a point in common then there exists and such that . The above system must have a unique solution. Let is verify this and find a common point.

Solution.

var('s, t')

Lt=(4+3*t, 3+t, 7+4*t)

Ls=(3-2*s, 2*s, 2+s)

sol=solve([Lt[0]==Ls[0],Lt[1]==Ls[1],Lt[2]==Ls[2]], s, t, solution_dict=True)

show(sol)

Clearly, gives a unique solution and point is the common point.

def axes(xmin=-1,xmax=1,ymin=-1, ymax=1, zmin=-1,zmax=1,**kwds):

ex = vector((1,0,0))

ey = vector((0,1,0))

ez = vector((0,0,1))

labels=text3d('x', (xmax+0.2,0,0)) + text3d("y", (0, ymax+0.20, 0)) + text3d("z", (0, 0, zmax+0.2))

return G+labels

A=axes(xmin=-3, xmax=3, ymin=-3, ymax=3, zmin=-3, zmax=3, color='red', thickness=2)

A

Cylinder and Traces

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given line segment, the axis of the cylinder. The solid enclosed by this surface and by two planes perpendicular to the axis is also called a cylinder. The surface area and the volume of a cylinder have been known since deep antiquity.

The cylinder is parallel to one of the coordinate axes.

DEFINITION 1  Cylinder

Given a curve in a plane and a line not in , a cylinder is the surface consisting of all lines parallel to given line that passes through a given plane curve .

DEFINITION 2  Trace

A trace of a surface is the set of points at which the surface intersects a plane that is parallel to one of the coordinate planes. The traces in the coordinate planes are called -trace, -trace and -trace.

Example 1

Draw a graph of the surface .

Figure 1  The surface is a parabolic cylinder.

Solution. Since there are no ’s in the equation, the trace of the graph in the plane is the same for every . The trace of the cylinder in every plane parallel to the -plane is parabola in the -plane and moving it in the direction of the -axis. The graph is a surface, called a parabolic cylinder and it is made up of infinitely many shifted copies of the same parabola. Here the rulings of the cylinder are parallel to the -axis.

var('x, y, z, v, u')

S=implicit_plot3d(x==z^2, (x,-2,2), (y,-2,2), (z,-2,2), plot_points=100, smooth=True, color='goldenrod', opacity=0.3)

p1=sum([parametric_plot3d((i^2 , v, i), (v,-2,2), color='red') for i in srange(-sqrt(2), sqrt(2), 0.5)])

p2=sum([parametric_plot3d((u^2 ,i, u), (u,- sqrt(2) , sqrt(2)), color='red') for i in srange(-2,2,0.5)])

show(S+p1+p2, frame=False)

■

Example 2

Sketch the graph of the surfaces in .

Solution. An equation in , is missing the variable . For all real values , the graph is a cylinder consisting of lies parallel to the -axis passing through the curve in the -plane.

Graph the curve in the -plane, which in the -trace of the surface.

Draw a second trace in a plane parallel to the -plane.

Draw lines parallel to the -axis passing through the two traces.

The result is a cylinder, running parallel to the -axis, consisting of copies of the curve .

Figure 2                    Figure 3

var('i,u,x,y,z')

p1=parametric_plot3d((i, 0, sin(i)), (i, -5, 5), opacity=0.5) + parametric_ plot3d((i, -5, sin(i)), (i, -5, 5), opacity=0.5)+parametric_plot3d((i, 5, sin(i)), (i, -5, 5), opacity=0.5)

p2=sum([parametric_plot3d((i, u, sin(i)), (u, - 5 ,  5), color=' red ') for i in srange(-5, 5, 0.5)])

show(p1+p2, frame=False)

var('x,y,z')

p =implicit_plot3d(z==cos(x), (x, -5, 5), (y, -5, 5), (z, -5, 5),opacity=0.5, smooth=True, axes=True) + text3d('z', (0,0,5), color=(0.5,0,0))+text3d("y", (0,5,0), color=(0,0.5,0))+text3d('x', (5,0,0), color=(0,0.5,0))

show(p, frame=False)

The general second-degree equation in ,

is a quadric surface. Here are constants. This equation can be brought into one of the two standard forms

or .

Quadric surfaces are the three dimensional counterparts of the conic sections in the plane.

Example 3

Sketch the quadric surface defined by the equation

.

This is an ellipsoid.

Solution. By substituting , we find that the trace in the -plane is which we recognize as an equation of an ellipse. In general, the horizontal trace in the plane is

which is an ellipse, provided that ＜9 that is, . The largest ellipse parallel to the -plane occurs with ; it is the -plane, which the ellipse

with axes of length and .

Similarly, the vertical traces are also ellipses:

,    ,   (if )

,    ,    (if ).

Figure 4  The ellipsoid

Figure 4 shows how drawing some traces indicates the shape of the surface. It’s called an ellipsoid because all of its traces are ellipses. Notice that it is symmetric with respect to each coordinate plane; this is a reflection of the fact that its equation involves only even powers of , and .

var('x,y,z')

p = implicit_plot3d((x^2)/25+(y^2)/16+(z^2)/9==1, (x, -10, 10), (y, -10, 10), (z, -10, 10), opacity=0.5, smooth=True, axes=True)+text3d("(0, 0, 3)", (0,0,3),  color=(0.5,0,0)) + t ext3d("(0, 4, 0)", (0,4,0), color=(0,0.5,0)) + text3d('z',  (0,0,10), color=(0.5,0,0)) + text3d("y", (0,10,0),  color = (0,0.5,0)) + text3d('x', (10, 0, 0), color=(0, 0.5, 0))

show(p, frame=False)

Example 4

Sketch the quadric surface defined by the equation

.

This is an ellipsoid paraboloid.

Figure 5

Solution. If we put , we get , so the -plane intersects the surface in a parabola. If we put (a constant), we get . This means that if we slice the graph with any plane parallel to the -plane, we obtain a parabola that opens leftward. Similarly, if , the trace is, which is again a parabola that opens leftward. If we put , we get the horizontal traces , which we recognize as a family of ellipses. Knowing the shapes of the traces, we can sketch the graph in Figure 5. Because of the elliptical and parabolic traces, the quadric surface is called an elliptic paraboloid.

var('x,y,z')

p =implicit_plot3d((x^2)*5+y^2==z, (x, -5, 5), (y, -5, 5),(z, -5, 5),opacity=0.5,smooth=True, axes=True)+text3d('z', (0,0,5), color=(0.5,0,0))+text3d("y", (0,5,0), color=(0,0.5,0))+text3d('x', (5,0,0), color=(0,0.5,0))

show(p,frame=False)

Example 5

Sketch the surface .

Solution. The traces in the vertical planes are the parabolas which open upward. The traces in are the parabolas , which open downward. The horizontal traces are , a family of hyperbolas. We draw the families of traces in Figure 6, and we show how the traces appear when placed in their correct planes in Figure 7.

Figure 6 Vertical traces are parabolas; horizontal traces are hyperbolas. All traces are labeled with the value of .

Figure 7  Traces moved to their correct planes.

In Figure 8, we fit together the traces from Figure 7 to form the surface a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle.

Figure 8 The surface is a hyperbolic paraboloid.

Example 6

Sketch the surface .

Solution. The trace in any horizontal plane is the ellipse

but the traces in the - and -planes are the hyperbolas

and  .

This surface is called a hyperboloid of one sheet and is sketched in Figure 9.

Figure 9

var('x,y,z')

p =implicit_plot3d((x^2)/9-y^2+(z^2)/9==1, (x, -10, 10), (y, -10, 10),(z, -10, 10),opacity=0.5,smooth=True, axes=True)+text3d('z', (0,0,10),  color=(0.5,0,0))+ text3d("y", (0,10,0), color=(0,0.5,0))+text3d('x', (10,0,0), color=(0,0.5,0))

show(p,frame=False)

Table 1 contains graphs of the six basic types of quadric surfaces in standard form. All surfaces are symmetric with respect to the -axis. If a quadric surface is symmetric about a different axis, its equation changes accordingly.

Example 7

Draw a graph of the surface .

Solution. Dividing by , we first put the equation in standard form:

.

Comparing this equation with Table 1, we see that it represents a hyperboloid of two sheets, the only difference being that in this case the axis of the hyperboloid is the -axis. The traces in the - and -planes are the hyperbolas

and    .

The surface has no trace in the -plane, but traces in the vertical planes for are the ellipses

,    .

which can be written as

,    .

These traces are used to make the sketch in Figure 10.

Figure 10

var('x,y,z')

p =implicit_plot3d((x^2)*5-y^2+2*z^2+5==0, (x, -5, 5), (y, -5, 5), (z, -5, 5), opacity=0.5, smooth=True, axes = True) + text3d('z', (0,0,5), color = (0.5,0,0)) + text3d("y", (0,5,0), color=(0,0.5,0)) + text3d('x', (5,0,0), color=(0, 0.5, 0)) + text3d("(0, -2, 0)", (0,-2,0), color = (0.5,0,0)) + text3d("(0, 2, 0)", (0, 2, 0), color=(0, 0.5, 0))

show(p,frame=False)

 Surface Equation & Traces Surface Equation & Traces Ellipsoid All traces are ellipses. If , the ellipsoid is a sphere. Cone Horizontal traces are ellip-ses. Vertical traces in the planes and are hyper-bolas if but are pairs of lines if . Elliptic Paraboloid Horizontal traces are ellip-ses. Vertical traces are para-bolas. The variable raised to the first power indicates the axis of the paraboloid. Hyperboloid of One Sheet Horizontal traces are ellip-ses. Vertical traces are hyper-bolas. The axis of symmetry cor-responds to the variable whose coefficient is nega-tive. Hyperbolic Paraboloid Horizontal traces are hy-per-bolas. Vertical traces are para-bolas. The case where is illustrated. Hyperboloid of Two Sheets Horizontal traces in are ellipses if or . Vertical traces are hyper-bolas. The two minus signs indicate two sheets.

TABLE 1 Graphs of quadric surfaces

Example 8

Solution. By completing the square we rewrite the equation as

Comparing this equation with Table 1, we see that it represents an elliptic paraboloid. Here, however, the axis of the paraboloid is parallel to the -axis, and it has been shifted so that its vertex is the point . The traces in the plane are the ellipses

,    .

The trace in the -plane is the parabola with equation , . The paraboloid is sketched in Figure 11.

Figure 11

var('x,y,z')

p = implicit_plot3d(x^2+2*y^2-2*x-z+4 ==0, (x, -5, 5), (y, -5, 5), (z, -5, 5), opacity=0.5, axes=True) + text3d('z',  (0, 0, 5), color=(0.5, 0, 0)) + text3d("y", (0, 5, 0),  color=(0, 0.5, 0)) + text3d('x', (5, 0, 0), color=(0, 0.5, 0)) + text3d("(1, 0, 3)", (1, 0, 3), color=(0, 0.5, 0))

show(p, frame=False)

Example 9

Sketch the equation

.

Figure 12

Solution.

var('x, y, z')

implicit_plot3d(x^2-3*y^2-2*z^2==0, (x, -1.5, 1.5), (y, -1.5, 1.5), (z, -1.5, 1.5), color='red', plot_points=80, smooth=True)

11.6 EXERCISES (Cylinders and Quadric Surfaces)

http://matrix.skku.ac.kr/Cal-Book/part2/CS-Sec-11-6-Sol.html

1.(a) What does the equation represent as a curve in ?

(b) What does it represent as a surface in ?

(c) What does the equation represent?

http://matrix.skku.ac.kr/cal-lab/cal-11-6-Exs-1.html

Solution. (a) Equation represents a parabola of slope passing through origin in .

(b) The equation of the graph is , which doesn't involve in . This means that any vertical plane with equation (parallel to the -plane) intersects the graph in a curve with , that is, a parabola. Below figure shows how the graph is formed by taking the parabola in the -plane and moving it in the direction of the -axis. So the graph is a surface, called a parabolic cylinder, made up of infinitely many shifted copies of the same parabola.

S=implicit_plot3d(z==y^2,(x,-2,2),(y,-2,2),(z,-2,2),plot_points=100,smooth=True,color='goldenrod', opacity=0.3)

p1=sum([parametric_plot3d([i,y,y^2],(y,-sqrt(2),sqrt(2)),color='red') for i in srange(-2,2,0.5)])

show(S+A+p1,frame=False)

(c) also represents a parabolic cylinder, this time with axis the -axis in .

2.(a) Sketch the graph of as a curve in .

(b) Sketch the graph of as a surface in .

(c) Describe and sketch the surface .

Solution. (a)

plot(exp(2*x), (x, -1, 1))

(b)

var('x, y, z')

implicit_plot3d(y==exp(2*x), (x, -1, 1), (y, -1, 1), (z, -1, 1))

(c)

var('x, y, z')

implicit_plot3d(z==exp(2*y), (x, -1, 1), (y, -1, 1), (z, -1, 1))

3-5. Describe and sketch the surface.

3.

Solution.

var('x, y, z')

implicit_plot3d(y^2+5*z^2-5==0, (x, -3, 3), (y, -3, 3), (z, -3, 3), opacity=0.5)

4.

Solution.

var('x, y')

z=cos(y)

plot3d(z, (x, -2, 2), (y, -pi, pi))

5.

Solution.

var('x, y, z')

implicit_plot3d(y*z==3, (x, -3, 3), (y, -10, 10), (z, -10, 10), opacity=0.5)

6-10. Find the traces of the given surface in , , . Then, identify the surface and sketch it.

6.

Solution.

var('x, y, z')

implicit_plot3d(x^2+9*y^2+9*z^2==9, (x, -3, 3), (y, -3, 3), (z, -3, 3), opacity=0.5)

7.

Solution. The trace in are ellipses of the form , , the trace in are parabolas of the form , and the trace in are parabolas of the form .

Combining these traces we form the graph.

8.

Solution.

var('x, y, z')

implicit_plot3d(25*x^2+z^2==100+4*y^2, (x, -3, 3), (y, -3, 3), (z, -3, 3),  opacity=0.5)

9.

Solution. The trace in are hyperbolas of the form , the trace in are circles of the form , , and the trace in are hyperbolas of the form .

Combining these traces we form the graph.

10.

Solution.

var('x, y, z')

implicit_plot3d(y^2+4*z^2-x==0, (x, -3, 3), (y, -3, 3), (z, -3, 3), opacity=0.5)

11-14. Reduce the equation to one of the standard forms, classify the surface, and sketch it.

11.

Solution. Dividing both sides by 15 gives , an elliptic paraboloid with vertex and axis the horizontal line .

12.

Solution.

var('x, y, z')

implicit_plot3d(z^2 == 9*x^2+4*y^2-5, (x, -3, 3), (y, -3, 3), (z, -3, 3), opacity=0.5)

13.

Solution. Completing squares in and gives

or ,

a hyperboloid of one sheet.

14.

Solution.

var('x, y, z')

implicit_plot3d(x^2 == 3*y^2+2*z^2, (x, -3, 3), (y, -3, 3), (z, -3, 3),  opacity=0.5)

15.

Solution.

var('x, y, z')

implicit_plot3d((x^2+9/4*y^2+z^2-1)^3-x^2*z^3-9/80 *y^2*z^3==0, (x, -1.5, 1.5), (y, -1.5, 1.5), (z, -1.5, 1.5), color='red', plot_points=80, smooth=True).show()

16.Sketch the region bounded by the surfaces and for .

Solution.

var('x, y, z')

z=(x^2+y^2)^(1/2)

plot3d(z, (x, -2, 2), (y, -2, 2), (z, 2, 4))

17. Find an equation for the surface obtained by rotating the parabola about the -axis. (Use revolution_plot3d to get the plot of this surface.)

18. Find an equation for the surface consisting of all points for which the distance from to the -axis is twice the distance from to the -plane. Identify the surface.

Solution. Let be an arbitrary point whose distance from th -axis is twice its distance from the -plane. The distance from to the -axis is and the distance from to the -plane() is .

Thus         .

So, the surface is a right circular cone with vertex the origin and axis the -axis.

19. Find an equation for the surface consisting of all points that are equidistant from the point and the plane

Calculus