Calculus-Sec-11-5-Solution

11.5  Equations of straight Lines and Planes

by SGLee - HSKim - JHLee

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

1. Through the point  and parallel to the vector .

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 .

(You may do it with Sage in http://math1.skku.ac.kr/ .

Open resources in http://math1.skku.ac.kr/pub/ )

parametric equation: ,

symmetric equation: .

3. Through the origin and parallel to the line .

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 .

Normal vector :  (2, -1, -2)

(ℝ)  =>

vector equation: ,

parametric equation: ,

symmetric equation: .

5. Through the origin and the point .

For this line, we have  and . Hence a vector equation is

and parametric equations are .

The symmetric equations are .

6. Through the points  and .

parametric equation: ,

symmetric equation: .

7. Through  and perpendicular to both  and .

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 ?

The lines are not parallel because the corresponding vectors

are not parallel.

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

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. ,

.

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

11. ℝ.

ℝ.

Since the direction vectors are  and ,

we have . Hence the lines are parallel.

12.  ,

.

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.  ,

.

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

14-15. Find an equation of the plane.

14. Through the point  and perpendicular to the vector .

.

15.Through the point  and with normal vector .

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?

,   ,

,  .

and  are parallel.

17. Which of the following four lines are parallel?

,    ,

,             .

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.

is a normal vector to the plane and  is a point of the plane.

Then  or  to be the equation of the plane.

19.

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.

The normal vector to the plane is n= <2, -1, 3>

21.

The distance .

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

22.

. Note that (-1, 0, 1) 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.

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.

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.

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