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Essential Calculus: Early Transcendentals
Found in: Page 565
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

(a) Find all vectors \({\bf{v}}\) such that

\(\langle 1,2,1\rangle \times {\bf{v}} = \langle 3,1, - 5\rangle \)

(b) Explain why there is no vector \({\bf{v}}\) such that

\(\langle 1,2,1\rangle \times {\bf{v}} = \langle 3,1,5\rangle \)

(a) The vectors \({\bf{v}}\) is \(\langle a,2a - 5,a - 1\rangle \).

(b) There exists no vectors of \({\bf{v}}\) that satisfies the equation\(\langle 1,2,1\rangle \times v = \langle 3,1,5\rangle \).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Formula used to express cross product between vector \(a\) and \(b\)

The expression for cross product between \(a\) and \(b\) vectors is,

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

Step 2: Calculation to find the vector \({\bf{v}}\)

(a)

It is given that,

\(\langle 1,2,1\rangle \times v = \langle 3,1,5\rangle \) …… (1)

Let \({\rm{v}} = \left\langle {{v_1},{v_2},{v_3}} \right\rangle \).

Compute \(\langle 1,2,1\rangle \times v\).

\(\begin{array}{l}\langle 1,2,1\rangle \times v = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&2&1\\{{v_1}}&{{v_2}}&{{v_3}}\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}2&1\\{{v_2}}&{{v_3}}\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}1&1\\{{v_1}}&{{v_3}}\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}1&2\\{{v_1}}&{{v_2}}\end{array}} \right|{\rm{k}}\\ = \left( {2{v_3} - {v_2}} \right){\rm{i}} - \left( {{v_3} - {v_1}} \right){\rm{j}} + \left( {{v_2} - 2{v_1}} \right){\rm{k}}\\ = \left\langle {\left( {2{v_3} - {v_2}} \right),\left( {{v_1} - {v_3}} \right),\left( {{v_2} - 2{v_1}} \right)} \right\rangle \end{array}\)

Substitute \(\langle 3,1, - 5\rangle \) for \(\langle 1,2,1\rangle \times v\),

\(\langle 3,1, - 5\rangle = \left\langle {\left( {2{v_3} - {v_2}} \right),\left( {{v_1} - {v_3}} \right),\left( {{v_2} - 2{v_1}} \right)} \right\rangle \)

Compare the R.H.S (Right hand side) and L.H.S (Left hand side).

\(2{v_3} - {v_2} = 3\) …… (2)

\(2{v_3} - {v_2} = 3\) …… (3)

\({v_2} - 2{v_1} = - 5\) …… (4)

Rearrange equation (4).

\({v_2} = 2{v_1} - 5\) …… (5)

Rearrange equation (3).

\({v_3} = {v_1} - 1\) …… (6)

Substitute \({v_1} - 1\) for \({v_3}\) and \(2{v_1} - 5\) for \({v_2}\) in equation (2),

\(\begin{array}{l}2\left( {{v_1} - 1} \right) - \left( {2{v_1} - 5} \right) = 3\\2{v_1} - 2 - 2{v_1} + 5 = 3\\ - 2 + 5 = 3\\3 = 3\end{array}\)

The vectors of \({\bf{v}}\) is in a dependent system.

Let \({v_1} = a\).

Substitute \(a\) for \({v_1}\) in equation (5) and in equation (6),

\(\begin{array}{l}{v_2} = 2a - 5\\{v_3} = a - 1\end{array}\)

The vectors of \({\rm{v}}\) is \(\langle a,2a - 5,a - 1\rangle \).

Thus, the vectors of \({\bf{v}}\) is \(\langle a,2a - 5,a - 1\rangle \).

Step 3: Calculation to prove that there is no vector \({\bf{v}}\) such that\(\langle 1,2,1\rangle   \times {\bf{v}} = \langle 3,1,5\rangle \)

(b)

Let \({\rm{v}} = \left\langle {{v_1},{v_2},{v_3}} \right\rangle \).

Compute \(\langle 1,2,1\rangle \times v\).

\(\begin{array}{l}\langle 1,2,1\rangle \times v = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&2&1\\{{v_1}}&{{v_2}}&{{v_3}}\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}2&1\\{{v_2}}&{{v_3}}\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}1&1\\{{v_1}}&{{v_3}}\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}1&2\\{{v_1}}&{{v_2}}\end{array}} \right|{\rm{k}}\\ = \left( {2{v_3} - {v_2}} \right){\rm{i}} - \left( {{v_3} - {v_1}} \right){\rm{j}} + \left( {{v_2} - 2{v_1}} \right){\rm{k}}\\ = \left\langle {\left( {2{v_3} - {v_2}} \right),\left( {{v_1} - {v_3}} \right),\left( {{v_2} - 2{v_1}} \right)} \right\rangle \end{array}\)

Substitute \(\langle 3,1,5\rangle \) for \(\langle 1,2,1\rangle \times {\rm{v}}\).

\(\langle 3,1, - 5\rangle = \left\langle {\left( {2{v_3} - {v_2}} \right),\left( {{v_1} - {v_3}} \right),\left( {{v_2} - 2{v_1}} \right)} \right\rangle \)

Compare the R.H.S (Right hand side) and the L.H.S (Left hand side) of the above equation.

\(2{v_3} - {v_2} = 3\) …… (7)

\({v_1} - {v_3} = 1\) …… (8)

\({v_2} - 2{v_1} = 5\) …… (9)

Rearrange equation (9).

\({v_2} = 2{v_1} + 5\) …… (10)

Rearrange equation (8).

\({v_3} = {v_1} - 1(11)\)

Substitute \({v_1} - 1\) for \({v_3}\) and \(2{v_1} + 5\) for \({v_2}\) in equation (7),

\(\begin{array}{l}2\left( {{v_1} - 1} \right) - \left( {2{v_1} + 5} \right) = 3\\2{v_1} - 2 - 2{v_1} - 5 = 3\\ - 2 - 5 = 3\\ - 7 = 3\end{array}\)

As LHS and RHS of the equation is not equal, it is an inconsistent system and has no solutions.

Hence, there exists no vectors of \({\bf{v}}\) that satisfies the equation\(\langle 1,2,1\rangle \times v = \langle 3,1,5\rangle \).

Most popular questions for Math Textbooks

To determine A geometric argument to show the vector \({\bf{c}} = s{\bf{a}} + t{\bf{b}}\).

(a) To find the parallel unit vectors to the tangent line of \(y = 2\sin x\).

(b) To find the perpendicular unit vectors to the tangent line of \(y = 2\sin x\).

(c) To sketch curve of \(y = 2\sin x\) along with vectors \( \pm \frac{1}{2}({\bf{i}} + \sqrt 3 {\bf{j}})\) and \( \pm \frac{1}{2}(\sqrt 3 {\bf{i}} - {\bf{j}})\).

To find the values of \(x\).

(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 \(R( - 1,4,7)\).

To show

(a) \({\rm{i}} \cdot {\rm{j}} = 0,{\rm{j}} \cdot {\rm{k}} = 0\) and \({\rm{k}} \cdot {\rm{i}} = 0\).

(b) \({\rm{i}} \cdot {\rm{i}} = 1,{\rm{j}} \cdot {\rm{j}} = 1\) and \({\rm{k}} \cdot {\rm{k}} = 1\)
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