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Q44E

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Found in: Page 565

### Essential Calculus: Early Transcendentals

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

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# (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 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$$\langle 1,2,1\rangle \times v = \langle 3,1,5\rangle$$.

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