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Expert-verified Found in: Page 565 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # (a) Let $$P$$ be a point not on the line $$L$$ that passes through the points $$Q$$ and $$R$$. Show that the distance $$d$$ from the point $$P$$ to the line $$L$$ is$$d = \frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}|}}$$where $${\bf{a}} = \overrightarrow {QR}$$ and $${\bf{b}} = \overrightarrow {QP}$$.(b) Use the formula in part (a) to find the distance from the point $$P(1,1,1)$$ to the line through $$Q(0,6,8)$$ and $$R( - 1,4,7)$$.

(a) The distance $$d$$ from the point $$P$$ to the line $$L$$ is $$d = \frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}|}}$$

(b) The distance $$(d)$$ from the point $$P(1,1,1)$$ to the line through the $$Q(0,6,8)$$ and $$R( - 1,4,7)$$ is $$\sqrt {\frac{{97}}{3}}$$.

See the step by step solution

## Step 1: Formula to find $$|{\bf{a}} \times {\bf{b}}|$$ in terms of $$\theta$$

(a)

The equation to find $$|{\bf{a}} \times {\bf{b}}|$$ in terms of $$\theta$$.

$$|{\bf{a}} \times {\bf{b}}| = |{\bf{a}}||{\bf{b}}|\sin \theta$$

Rearrange the equation.

$$\sin \theta = \frac{{|{\bf{a}}||{\bf{b}}|}}{{|{\bf{a}} \times {\bf{b}}|}}$$

## Step 2: Calculation to show that the distance $$d$$ from the point $$P$$ to the line $$L$$ is $$d = \frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}|}}$$

It is given that,

$${\bf{a}} = \overrightarrow {QR}$$ and $${\bf{b}} = \overrightarrow {QP}$$

Let $$P$$ be a point not on the line $$L$$ that passes through the points $$Q$$ and $$R$$ as shown in Figure 1. Here,

$$|\overrightarrow {PS} | = d$$ that is, the perpendicular length from the point to line.

From triangle $$PQS$$, write the expression for $$\sin \theta$$.

$$\sin \theta = \frac{{|\overrightarrow {PS} |}}{{|\overrightarrow {QP} |}}$$

Substitute $$d$$ for $$|\overrightarrow {PS} |$$ and $$|{\bf{b}}|$$ for $$|\overrightarrow {QP} |$$,

$$\sin \theta = \frac{d}{{|{\bf{b}}|}}$$

Rearrange the equation.

$$d = \sin \theta |{\bf{b}}|$$

Substitute $$\frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}||{\bf{b}}|}}$$ for $$\sin \theta$$,

$$\begin{array}{l}d = \left( {\frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}||{\bf{b}}|}}} \right)|{\bf{b}}|\\ = \frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}|}}\end{array}$$

Thus, the distance $$d$$ from the point $$P$$ to the line $$L$$ is $$d = \frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}|}}$$.

## Step 3: Formula to express the cross product between the vectors $${\bf{a}}$$ and $${\bf{b}}$$

(b)

The equation for the cross product between the vectors $${\bf{a}}$$ and $${\bf{b}}$$ is,

$${\bf{a}} \times {\bf{b}} = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|$$ …… (1)

The magnitude of a three-dimensional vector $${\bf{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle$$ is,

$$|{\bf{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2} {\rm{ }}$$ …… (2)

The equation for distance $$d$$ from the point $$P$$ to the line $$L$$ is,

$$d = \frac{{|{\bf{a}} \times {\bf{b}}|}}{{|{\bf{a}}|}}{\rm{ }}$$ …… (3)

## Step 4: Calculation to find the distance from the point $$P(1,1,1)$$ to the line through $$Q(0,6,8)$$ and $$R( - 1,4,7)$$

Find $${\bf{a}} = \overrightarrow {QR}$$.

$$\begin{array}{l}{\bf{a}} = \overrightarrow {QR} \\ = R( - 1,4,7) - Q(0,6,8)\\ = \langle ( - 1,0),(4, - 6),(7, - 8)\rangle \\ = \langle - 1, - 2, - 1\rangle \end{array}$$

Find $${\bf{b}} = \overrightarrow {QP}$$.

$$\begin{array}{l}{\bf{b}} = \overrightarrow {QP} \\ = P(1,1,1) - Q(0,6,8)\\ = \langle ( - 1,0),(1, - 6),(1, - 8)\rangle \\ = \langle 1, - 5,7\rangle \end{array}$$

Substitute $$- 1$$ for $${a_1}$$, $$- 2$$ for $${a_2}$$, $$- 3$$ for $${a_3}$$, $$1$$ for $${b_1}$$, $$- 5$$ for $${b_2}$$ and $$- 7$$ for $${b_3}$$ in equation (1),

$$\begin{array}{l}{\bf{a}} \times {\bf{b}} = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{ - 1}&{ - 2}&{ - 1}\\1&{ - 5}&{ - 7}\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}{ - 2}&{ - 1}\\{ - 5}&{ - 7}\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}{ - 1}&{ - 1}\\1&{ - 7}\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\1&{ - 5}\end{array}} \right|{\bf{k}}\\ = (14 - 5){\bf{i}} - (7 + 1){\bf{j}} + (5 + 2){\bf{k}}\\ = 9{\bf{i}} - 8{\bf{j}} + 7{\bf{k}}\end{array}$$

$${\bf{a}} \times {\bf{b}} = \langle 9, - 8,7\rangle$$

Substitute $$- 1$$ for $${a_1}$$, $$- 2$$ for $${a_2}$$ and $$- 1$$ for $${a_3}$$ in equation (2),

$$\begin{array}{l}|{\bf{a}}| = \sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{( - 1)}^2}} \\ = \sqrt {1 + 4 + 1} \\ = \sqrt 6 \end{array}$$

Substitute $$\langle 9, - 8,7\rangle$$ for $${\bf{a}} \times {\bf{b}}$$ and $$\sqrt 6$$ for $$|{\bf{a}}|$$ in equation (3),

$$\begin{array}{l}d = \frac{{|9, - 8,7\rangle \mid }}{{\sqrt 6 }}\\ = \frac{{\sqrt {{{(9)}^2} + {{( - 8)}^2} + {{(7)}^2}} }}{{\sqrt 6 }}\\ = \sqrt {\frac{{81 + 64 + 49}}{6}} \\ = \sqrt {\frac{{194}}{6}} \\d = \sqrt {\frac{{97}}{3}} \end{array}$$

The distance $$(d)$$ from the point $$P(1,1,1)$$ to the line through the points $$Q(0,6,8)$$ and $$R( - 1,4,7)$$ is $$\sqrt {\frac{{97}}{3}}$$. ### Want to see more solutions like these? 