Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

(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 \(\sqrt {\frac{{97}}{3}} \).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 \(\sqrt {\frac{{97}}{3}} \).

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Essential Calculus: Early Transcendentals

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Short Answer

Step by Step Solution

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

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

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

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

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Step by Step Solution

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

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

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

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

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