Solutions to Problem Set 6

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.

$$x = 5 s, \quad y = 6 - 2 s, \quad z = 3 + s.$$

$$x = 4 - 10 t, \quad y = 1 + 4 t, \quad z = 5 - 2 t.$$

The first line is parallel to the vector $(5, -2, 1)$ . The second line is parallel to the vector $(-10, 4, -2)$ . The vectors are multiples:

$$(-2) \cdot (5, -2, 1) = (-10, 4, -2).$$

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.

$$x = 3 + s, \quad y = 2 - 5 s, \quad z = 1 + 7 s.$$

$$x = -2 + 3 t, \quad y = -5 + t, \quad z = 4 t.$$

The first line is parallel to the vector $(1, -5, 7)$ . The second line is parallel to the vector $(3, 1, 4)$ . The vectors parallel to the lines are not multiples, so I see if the lines intersect. Equate the x-expressions:

$$\eqalign{ 3 + s & = -2 + 3 t \cr s & = -5 + 3 t \cr}$$

Equate the y-expressions and substitute $s = -5 + 3 t$ :

$$\eqalign{ 2 - 5 s & = -5 + t \cr 2 - 5(-5 + 3 t) & = -5 + t \cr 27 - 15 t & = -5 + t \cr 32 & = 16 t \cr 2 & = t \cr}$$

This gives $s = -5 + 3 \cdot 2 =
   1$ . Plug these into the z-equations to see if they are consistent:

$$z = 1 + 7 \cdot 1 = 8, \quad z = 4 \cdot 2 = 8.$$

Since the values agree, the lines intersect. Plugging $s = 1$ into the first set of equations gives the intersection point $(4, -3, 8)$ .


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

$$x = 2 + 3 s, \quad y = 4 - s, \quad z = 1 + 6 s.$$

$$x = 5 + t, \quad y = 3 - 2 t, \quad z = 8 + t.$$

The first line is parallel to the vector $(3, -1, 6)$ . The second line is parallel to the vector $(1, -2, 1)$ . The vectors parallel to the lines are not multiples, so I see if the lines intersect. Equate the x-expressions:

$$\eqalign{ 2 + 3 s & = 5 + t \cr -3 + 3 s & = t \cr}$$

Equate the y-expressions and substitute $t = -3 + 3 s$ :

$$\eqalign{ 4 - s & = 3 - 2 t \cr 4 - s & = 3 - 2(-3 + 3 s) \cr 4 - s & = 9 - 6 s \cr 5 s & = 5 \cr s & = 1 \cr}$$

This gives $t = -3 + 3 \cdot 1 =
   6$ . Plug these into the z-equations to see if they are consistent:

$$z = 1 + 6 \cdot 1 = 7, \quad z = 8 + 6 = 14.$$

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 $(9, -3, 1)$ and is perpendicular to the line

$$x = 1 + 3 t, \quad y = 5 + 2 t, \quad z = -7 t.$$

The vector $(3, 2, -7)$ is parallel to the line. Since the plane I want is perpendicular to the line, the plane must be perpendicular to the vector $(3, 2, -7)$ . The plane is

$$3 (x - 9) + 2 (y + 3) - 7 (z - 1) = 0, \quad\hbox{or}\quad 3 x + 2 y - 7 z = 14.\quad\halmos$$


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

$$4 x - y + 2 z = 32 \quad\hbox{and}\quad \left\{\matrix{ x = 3 + t \cr y = 3 + 5 t \cr z = 1 + 4 t \cr}\right\}.$$

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

$$\eqalign{ 4 (3 + t) - (3 + 5 t) + 2 (1 + 4 t) & = 32 \cr 7 t + 11 & = 32 \cr 7 t & = 21 \cr t & = 3 \cr}$$

Substitute $t = 3$ into the line equations to find the intersection point:

$$x = 3 + 3 = 6, \quad y = 3 + 15 = 18, \quad z = 1 + 12 = 13.$$

The intersection point is $(6,
   18, 13)$ .


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

$$x + 5 y - 3 z = 2 \quad\hbox{and}\quad 2 x + y - 4 z = 7.$$

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:

$$\eqalign{ 2 x + 10 y - 6 z & = 4 \cr 2 x + y - 4 z & = 7 \cr \noalign{\vskip2pt \hrule \vskip2pt} 9 y - 2 z & = -3 \cr}$$

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 $9 - 12 = -3$ . That is,

$$9 \cdot 1 - 2 \cdot 6 = -3.$$

So $y = 1$ and $z = 6$ works. Plugging this into the first plane equation and solving for x, I get

$$\eqalign{ x + 5 \cdot 1 - 3 \cdot 6 & = 2 \cr x - 13 & = 2 \cr x & = 15 \cr}$$

Thus, $P(15, 1, 6)$ is a point on the line.

Going back to $9 y - 2 z = -3$ , I notice that $-9 + 6 = -3$ . That is,

$$9 \cdot (-1) - 2 \cdot (-3) = -3.$$

So $y = -1$ and $z = -3$ works. Plugging this into the first plane equation and solving for x, I get

$$\eqalign{ x + 5 \cdot (-1) - 3 \cdot (-3) & = 2 \cr x + 4 & = 2 \cr x & = -2 \cr}$$

Thus, $Q(-2, -1, -3)$ is another point on the line.

$$\hbox{\epsfysize=1.5in \epsffile{lines-and-planes-problems-3.eps}}$$

The vector $\bvec{P Q} = (-17,
   -2, -9)$ is parallel to the line. Using $(15, 1, 6)$ as the point on the line, the line is

$$x - 15 = -17 t, \quad y - 1 = -2 t, \quad z - 6 = -9 t.\quad\halmos$$

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 $P(1, 4, 3)$ and the line

$$x - 3 = 2 t, \quad y + 5 = t, \quad z - 8 = 3 t.$$

I need a point on the plane and a vector perpendicular to the plane. For the point, I can use $P(1, 4,
   3)$ .

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 $(2, 1, 3)$ is parallel to the line. Since the line lies in the plane, the vector $(2, 1, 3)$ does as well.

Next, setting $t = 0$ gives the point $Q(3, -5, 8)$ on the line. Hence, Q lies in the plane, and the vector $\bvec{P Q} = (2, -9, 5)$ does as well. Take the cross product:

$$(2, 1, 3) \times (2, -9, 5) = \left|\matrix{ \ihat & \jhat & \khat \cr 2 & 1 & 3 \cr 2 & -9 & 5 \cr}\right| = (32, -4, -20).$$

I can divide all the components by 4 without changing the vector's direction; I get $(8, -1, -5)$ . Using this vector and the point $(1, 4, 3)$ , the plane is

$$8 (x - 1) - (y - 4) - 5 (z - 3) = 0, \quad\hbox{or}\quad 8 x - y - 5 z = -11.\quad\halmos$$


8. The following lines intersect:

$$x = 1 + s, \quad y = 3 - 2 s, \quad z = 5 + 3 s.$$

$$x = -1 + t, \quad y = 3 + t, \quad z = 2 t.$$

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 $s = 0$ in the first line, I find that $(1, 3, 5)$ 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 $(1, -2, 3)$ is parallel to the first line, so it's parallel to the plane. The vector $(1, 1,
   2)$ is parallel to the second line, so it's parallel to the plane. Take their cross product:

$$(1, -2, 3) \times (1, 1, 2) = \left|\matrix{ \ihat & \jhat & \khat \cr 1 & -2 & 3 \cr 1 & 1 & 2 \cr}\right| = (-7, 1, 3).$$

The plane is

$$-7 (x - 1) + (y - 3) + 3 (z - 5) = 0, \quad\hbox{or}\quad -7 x + y + 3 z = 11.\quad\halmos$$


9. The following lines are parallel:

$$x = 7 - 2 s, \quad y = 3 + 4 s, \quad z = 10 s.$$

$$x = 5 + t, \quad y = 1 - 2 t, \quad z = 3 - 5 t.$$

Find the equation of the plane that contains them.

I need a point on the plane and a vector perpendicular to the plane. Setting $s = 0$ in the first line, I find that $P(7, 3, 0)$ 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 $(1, -2, -5)$ is parallel to the second line, so it's parallel to the plane.

$$\hbox{\epsfysize=1.5in \epsffile{lines-and-planes-problems-4.eps}}$$

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 $\vec{0}$ . Instead, set $t = 0$ in the second line. This gives the point $Q(5, 1, 3)$ in the second line. Since P and Q both lie in the plane, $\bvec{P Q} = (-2, -2,
   3)$ 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:

$$(1, -2, -5) \times (-2, -2, 3) = \left|\matrix{ \ihat & \jhat & \khat \cr 1 & -2 & -5 \cr -2 & -2 & 3 \cr}\right| = (-16, 7, -6).$$

The vector $(-16, 7, -6)$ is perpendicular to the plane. Using $P(7, 3, 0)$ as the point on the plane, the plane is

$$-16 (x - 7) + 7 (y - 3) - 6 z = 0, \quad\hbox{or}\quad -16 x + 7 y - 6 z = -91.\quad\halmos$$


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


Contact information

Bruce Ikenaga's Home Page

Copyright 2018 by Bruce Ikenaga