Solutions to Problem Set 7

Math 311-01/02

2-5-2018

1. Find the distance from the point $P(1, -2, 4)$ to the plane

$$3 x + y - 5 z = 10.$$

By juggling numbers, I can see that $Q(0, 0, -2)$ is a point on the plane. The vector $\bvec{P Q}
   = (-1, 2, -6)$ goes from P to Q.

The vector $\vec{v} = (3, 1, -5)$ is perpendicular to the plane. The (absolute value of the) component of $\bvec{P Q}$ in the direction of $\vec{v}$ gives the distance.

$$\hbox{\epsfysize=1.5in \epsffile{line-and-plane-distance-problems-1.eps}}$$

$$ \comp_{\vec{v}} \bvec{P Q} = \dfrac{\bvec{P Q} \cdot \vec{v}}{\|\vec{v}\|} = \dfrac{(-1, 2, -6) \cdot (3, 1, -5)}{\|(3, 1, -5)\|} = \dfrac{29}{\sqrt{35}}.$$

Since this is positive, I don't need to take absolute values. The distance is $\dfrac{29}{\sqrt{35}} = 4.90189 \ldots$ .


2. Find the distance from the point $P(1, 0, -3)$ to the line

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

Setting $t = 0$ , I see that $Q(1, 0, 2)$ is a point on the line. Therefore, $\bvec{P Q}
   = (0, 0, 5)$ . Reading off the coefficients of t's, I also see that $\vec v = (2, -1, 2)$ is parallel to the line.

Therefore,

$$\comp_{\vec{v}} \bvec{P Q} = \dfrac{\bvec{P Q} \cdot \vec{v}}{\|\vec{v}\|} = \dfrac{(0, 0, 5) \cdot (2, -1, 2)}{|(2, -1, 2)\|} = \dfrac{10}{3}.$$

$$\hbox{\epsfysize=1.5in \epsffile{line-and-plane-distance-problems-2.eps}}$$

The distance from P to the line is the leg of a right triangle, and I know the hypotenuse ($\|\bvec{P
   Q}\| = \sqrt{25} = 5$ ) and the other leg ($\comp_{\vec{v}}
   \bvec{P Q}$ ). Hence, I can find the distance using Pythagoras:

$$\hbox{distance} = \sqrt{\|\bvec{P Q}\|^2 - \left(\comp_{\vec{v}} \bvec{P Q}\right)^2} = \sqrt{25 - \dfrac{100}{9}} = \dfrac{5\sqrt{5}}{3} = 3.72677 \ldots.\quad\halmos$$


3. The following planes are parallel:

$$2 x - y + 5 z = 1 \quad\hbox{and}\quad -6 x + 3 y - 15 z = 12.$$

Find the distance between them.

This example uses a setup like the one for the distance from a point to a plane.

Setting $x = 0$ and $z
   = 0$ in the first plane gives $-y = 1$ , so $y = -1$ . The point $P(0, -1, 0)$ lies on the first plane.

Setting $y = 0$ and $z
   = 0$ in the second plane gives $-6 x = 12$ , so $x = -2$ . The point $Q(-2, 0, 0)$ lies on the second plane.

The vector $\bvec{P Q} = (-2, 1,
   0)$ goes from the first plane to the second plane.

The vector $(2, -1, 5)$ is perpendicular to the first plane (and also to the second plane, since the planes are parallel).

The distance is given by the absolute value of the component of $\bvec{P Q}$ on the perpendicular vector $(2, -1, 5)$ :

$$\comp_{(2, -1, 5)} \bvec{P Q} = \dfrac{(-2, 1, 0) \cdot (2, -1, 5)}{\|(2, -1, 5)\|} = \dfrac{-4 - 1 + 0}{\sqrt{30}} = -\dfrac{5}{\sqrt{30}}.$$

The distance is given by the absolute value: $\dfrac{5}{\sqrt{30}} = 0.91287
   \ldots$ .


4. The following lines are skew:

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

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

Find the distance between them.

This example uses a setup like the one for the distance from a point to a plane.

Skew lines lie in parallel planes: Think of one line as lying in the "ceiling" and the other line as lying in the "floor". So the method here is similar to the method for finding the distance between parallel planes.

Setting $s = 0$ gives the point $P(1, 3, -2)$ in the first line.

Setting $t = 0$ gives the point $Q(0, 4, 1)$ in the second line.

The vector $\bvec{P Q} = (-1, 1,
   3)$ goes from the first line to the second.

The vector $(1, 5, 3)$ is parallel to the first line. The vector $(-2, 1, 6)$ is parallel to the second line. Their cross product is a vector perpendicular to both lines:

$$(1, 5, 3) \times (-2, 1, 6) = \left|\matrix{ \ihat & \jhat & \khat \cr 1 & 5 & 3 \cr -2 & 1 & 6 \cr}\right| = (27, -12, 11).$$

$$\hbox{\epsfysize=1.5in \epsffile{line-and-plane-distance-problems-3.eps}}$$

The distance is given by the absolute value of the component of $\bvec{P Q}$ on the perpendicular vector $(27, -12, 11)$ :

$$\comp_{(27, -12, 11)} \bvec{P Q} = \dfrac{(-1, 1, 3) \cdot (27, -12, 11)}{\|(27, -12, 11)\|} = \dfrac{-27 - 12 + 33}{\sqrt{994}} = \dfrac{-6}{\sqrt{994}}.$$

The distance is given by the absolute value: $\dfrac{6}{\sqrt{994}} = 0.19030
   \ldots$ .


5. The following lines are parallel:

$$x = 1 - 2 s, \quad y = 3 + 4 s, \quad z = 10s.$$

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

Find the distance between them.

This example uses a setup like the one for the distance from a point to a line.

Setting $s = 0$ gives the point $P(1, 3, 0)$ on the first line.

Setting $t = 0$ gives the point $Q(3, 5, 4)$ on the second line.

Hence, $\bvec{P Q} = (2, 2, 4)$ is a vector from the first line to the second line.

The vector $(1, -2, -5)$ is parallel to the second line (and also to the first line, since the lines are parallel. Then

$$\comp_{(1, -2, -5)} \bvec{P Q} = \dfrac{(2, 2, 4) \cdot (1, -2, -5)}{\|(1, -2, -5)\|} = \dfrac{2 - 4 - 20}{\sqrt{30}} = -\dfrac{22}{\sqrt{30}}.$$

$$\hbox{\epsfysize=1.5in \epsffile{line-and-plane-distance-problems-4.eps}}$$

The distance is given by Pythagoras:

$$\sqrt{\|\bvec{P Q}\|^2 - \left(\comp_{(1, -2, -5)} \bvec{P Q}\right)^2} = \sqrt{24 - \dfrac{484}{30}} = \sqrt{\dfrac{236}{30}} = \sqrt{\dfrac{118}{15}} = 2.80475 \ldots.\quad\halmos$$


6. (a) The following line and plane are parallel:

$$x = 3 t, \quad y = 1 + 4 t, \quad z = 5 + t.$$

$$x + 2 y - 11 z = 3.$$

Explain why they are parallel.

(b) Find the distance between them.

(a) The vector $(3, 4, 1)$ is parallel to the line. The vector $(1, 2, -11)$ is perpendicular to the plane.

$$(3, 4, 1) \cdot (1, 2, -11) = 3 + 8 - 11 = 0.$$

The vectors are perpendicular, so the line is parallel to the plane.

(b) This example uses a setup like the one for the distance from a point to a plane.

Setting $t = 0$ , I find that the point $P(0, 1, 5)$ is on the line. Note that

$$1 + 2 \cdot 1 - 11 \cdot 0 = 3.$$

Therefore, the point $Q(1, 1,
   0)$ in on the plane. The vector $\bvec{P Q} = (1, 0, -5)$ goes from the line to the plane. The distance is given by the absolute value of the component of $\bvec{P Q}$ on the perpendicular vector $(1, 2, -11)$ :

$$\comp_{(1, 2, -11)} \bvec{P Q} = \dfrac{(1, 0, -5) \cdot (1, 2, -11)}{\|(1, 2, -11)\|} = \dfrac{1 + 0 + 55}{\sqrt{126}} = \dfrac{56}{\sqrt{126}}.$$

The distance is $\dfrac{56}{\sqrt{126}} = 4.98887 \ldots$ .


The first rule is to keep an untroubled spirit. The second is to look things in the face and know them for what they are. - Marcus Aurelius


Contact information

Bruce Ikenaga's Home Page

Copyright 2018 by Bruce Ikenaga