Math 311-01/02

2-2-2018

* [Lines and planes problems]*

1. Determine whether the following lines are parallel or skew or if they intersect. If they intersect, find their intersection point.

The first line is parallel to the vector . The second line is parallel to the vector . The vectors are multiples:

Hence, the vectors are parallel, and the lines are parallel.

2. Determine whether the following lines are parallel or skew or if they intersect. If they intersect, find their intersection point.

The first line is parallel to the vector . The second line is parallel to the vector . The vectors parallel to the lines are not multiples, so I see if the lines intersect. Equate the x-expressions:

Equate the y-expressions and substitute :

This gives . Plug these into the z-equations to see if they are consistent:

Since the values agree, the lines intersect. Plugging into the first set of equations gives the intersection point .

3. Determine whether the following lines are parallel or skew or if they intersect. If they intersect, find their intersection point.

The first line is parallel to the vector . The second line is parallel to the vector . The vectors parallel to the lines are not multiples, so I see if the lines intersect. Equate the x-expressions:

Equate the y-expressions and substitute :

This gives . Plug these into the z-equations to see if they are consistent:

Since the values don't agree, the lines don't intersect. Therefore, the lines are skew.

4. Find the equation of the plane which passes through the point and is perpendicular to the line

The vector is parallel to the line. Since the plane I want is perpendicular to the line, the plane must be perpendicular to the vector . The plane is

5. Find the point of intersection of the following plane and line:

Substitute the line expressions for x, y, and z into the plane equation and solve for t:

Substitute into the line equations to find the intersection point:

The intersection point is .

6. Find the parametric equation for the line of intersection of the planes

I need a point on the line and a vector parallel to the line. There are many ways to do this problem.

To find a point on the line, solve the plane equations simultaneously. Note that since I have 2 equations but 3 variables, I won't get a unique solution --- in fact, since the intersection is a line, and a line contains infinitely many points, there must be infinitely many solutions.

For example, I'll multiply the first equation by 2 and subtract the second equation to eliminate x:

At this point, any y and z satisfying this equation will do, and I can just "juggle numbers" to get a solution. For instance, I notice that . That is,

So and works. Plugging this into the first plane equation and solving for x, I get

Thus, is a point on the line.

Going back to , I notice that . That is,

So and works. Plugging this into the first plane equation and solving for x, I get

Thus, is another point on the line.

The vector is parallel to the line. Using as the point on the line, the line is

Note you could also find a vector parallel to the line by taking vectors perpendicular to each of the planes and taking their cross product.

7. Find the equation of the plane which contains the point and the line

I need a point on the plane and a vector perpendicular to the plane. For the point, I can use .

To get a vector perpendicular to the plane, I'll take the cross product of two vectors in (or parallel to) the plane. The vector is parallel to the line. Since the line lies in the plane, the vector does as well.

Next, setting gives the point on the line. Hence, Q lies in the plane, and the vector does as well. Take the cross product:

I can divide all the components by 4 without changing the vector's direction; I get . Using this vector and the point , the plane is

8. The following lines intersect:

Taking this for granted, find the equation of the plane that contains them.

I need a point on the plane and a vector perpendicular to the plane. Setting in the first line, I find that is on the line. Hence, it is a point on the plane.

For a vector perpendicular to the plane, I'll take the cross product of two vectors in (or parallel to) the plane. The vector is parallel to the first line, so it's parallel to the plane. The vector is parallel to the second line, so it's parallel to the plane. Take their cross product:

The plane is

9. The following lines are parallel:

Find the equation of the plane that contains them.

I need a point on the plane and a vector perpendicular to the plane. Setting in the first line, I find that is on the line. Hence, it is a point on the plane.

For a vector perpendicular to the plane, I'll take the cross product of two vectors in (or parallel to) the plane. The vector is parallel to the second line, so it's parallel to the plane.

For a second vector, I can't use a vector parallel to the first line, since such a vector will be parallel to the second line's vector, and their cross product will be . Instead, set in the second line. This gives the point in the second line. Since P and Q both lie in the plane, lies in the plane. Since it goes from one of the lines to the other, it can't be parallel to the lines. Take the cross product of the two vectors:

The vector is perpendicular to the plane. Using as the point on the plane, the plane is

*If you want an easy life, die.* - *Richard Platek*

Copyright 2018 by Bruce Ikenaga