Q.80 - Calculus

Question

The product$\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$ is an example of a vector called a vector triple product.

(a) Show that if $v = v 1, v 2, v 3$and $w = w1, w2, w3,$then $i \times (v \times w) = w1v - v 1w.$

(b) Derive similar expressions from $j \times (v \times w)$ and $k \times (v \times w).$

(c) Use your results from parts (a) and (b) to show that $u \times (v \times w) = (u · w)v - (u · v)w.$

(d) Use your results from part (c) and the anti commutativity of the cross product to derive a similar expression for the vector triple product $(u \times v) \times w.$

(e) Use your results from parts (c) and (d) to show that the cross product is not associative.

(f) Under what conditions is

                $u \times (v \times w) = (u \times v) \times w?$

Step-by-Step Solution

Answer

part (A)

a) Hence, the result is proved.

Part (B)

b) The value of $\mathbf{k}\times (\mathbf{v}\times \mathbf{w})\text{ is }{w}_3\mathbf{v}-v_3\mathbf{w}$ 

Part (C)

c) Hence, the result is proved.

Part (D)

d) The value of  $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}\text{ is }(\mathbf{u}·\mathbf{v})\mathbf{w}-(\mathbf{u}·\mathbf{w})\mathbf{v}$

Part (E)

e) Therefore, the cross product is not associative.

Part (F)

f) Thus, the condition for$\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$ to hold is that the vector $\mathbf{u}$ is orthogonal to both $\mathit{w}$ and $\mathit{v}$

Step 1 Introduction

(a) Take the vectors  and .

The goal is to demonstrate that $\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=w_1\mathbf{v}-v_1\mathbf{w}$.

Find the cross product $\mathrm{i}\times (\mathrm{v}\times \mathbf{w})$  to prove the result.

Step 2 Given Information

The by-product is $(\mathbf{v}\times \mathbf{W})$  stands for:

 $\mathbf{v}\times \mathbf{w}=$$\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right|$

$=\left(v_2{w}_3-v_3{w}_2\right)\mathbf{i}+\left(v_3{w}_1-v_1{w}_3\right)\mathbf{j}+\left(v_1{w}_2-v_2{w}_1\right)\mathbf{k}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$                          

             

Step 3 Explanation

The by-product  $\mathbf{i}\times (\mathbf{v}\times \mathbf{w})$ is

  $\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 0& 0\\ v_2{w}_3-v_3{w}_2& v_3{w}_1-v_1{w}_3& v_1{w}_2-v_2{w}_1\end{array}\right|\text{ (Using 1) }$

   $=0\mathbf{i}-\left(v_1{w}_2-v_2{w}_1\right)\mathbf{j}+\left(v_3{w}_1-v_1{w}_3\right)\mathbf{k}$

    

    

Step 4 Explanation

The expression$w_1\mathbf{v}-v_1\mathbf{w}$   yields:   

 

Step 5 Explanation

The following are the results of the equations (1) and (2)  $\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=w_1\mathbf{v}-v_1\mathbf{w}$

Step 6 Given Information

(b) Consider the vectors  and .

The goal is to calculate $\mathbf{j}\times (\mathbf{v}\times \mathbf{w})\text{ and }\mathbf{k}\times (\mathbf{v}\times \mathbf{w})$.

Use the result $\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=w_1\mathbf{v}-v_1\mathbf{w}$ to determine the value of  $\mathbf{j}\times (\mathbf{v}\times \mathbf{w})$ to get the value of $j\times (v\times w)$.

Step 7 Explanation

The value of $\mathbf{j}\times (\mathbf{v}\times \mathbf{w})$ in generalizing the result $\mathbf{i}\times (\mathbf{v}\times \mathbf{W})={\mathbf{w}}_1\mathbf{v}-v_1\mathbf{W}$ is:

$j\times (\mathbf{v}\times \mathbf{w})=w_2\mathbf{v}-v_2\mathbf{w}$ 

As a result , $\mathbf{j}\times (\mathbf{v}\times \mathbf{w})\text{ is }{w}_2\mathbf{v}-v_2\mathbf{w}$

Step 8 Explanation

The value of $\mathbf{k}\times (\mathbf{v}\times \mathbf{w})$ when generalising the result $\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=w_1\mathbf{v}-v_1\mathbf{w}$ is:

$\mathbf{k}\times (\mathbf{v}\times \mathbf{w})=w_3\mathbf{v}-v_3\mathbf{w}$ 

As a result, $\mathbf{k}\times (\mathbf{v}\times \mathbf{w})\text{ is }{\mathbf{w}}_3\mathbf{v}-v_3\mathbf{w}$

Step 9 Given Infomation

(c) The goal is to demonstrate that $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v})\mathbf{w}$.

Use the following relationships to show your point.

$\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=w_1\mathbf{v}-v_1\mathbf{w}$

$\mathbf{j}\times (\mathbf{v}\times \mathbf{w})=w_2\mathbf{v}-v_2\mathbf{w}$

$\mathbf{k}\times (\mathbf{v}\times \mathbf{w})={\mathbf{w}}_3\mathbf{v}-v_3\mathbf{w}$

Step 10 Explanation

The expression$\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=w_1\mathbf{v}-v_1\mathbf{w}$  can be expressed as follows: $\mathbf{i}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{i}·\mathbf{w})\mathbf{v}-(\mathbf{i}·\mathbf{v})\mathbf{w}$ (Because$\mathbf{i}·\mathbf{i}=1,\mathbf{i}·\mathbf{j}=0,\mathbf{i}·\mathbf{k}=0)$  As a result of generalising the result, the following equation emerges. $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v})\mathbf{w}$

The result is proven.

Step 11 Given Information

(d) The goal is to calculate the result of $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$

Use the result $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v})\mathbf{w}$ and the anti-commutativity of the vectors to get the result.

Step 12 Explanation

Due to the fact that the vectors are not commutative.

As a result, $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}=-\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$

Step 13 Explanation

As a result,$(\mathbf{u}\times \mathbf{v})\times \mathbf{w}=-\left[\right(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v}\left)\mathbf{w}\right]$ (Substitution) $=(\mathbf{u}·\mathbf{v})\mathbf{w}-(\mathbf{u}·\mathbf{w})\mathbf{v}$

As a result,  $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}\text{ is }(\mathbf{u}·\mathbf{v})\mathbf{w}-(\mathbf{u}·\mathbf{w})\mathbf{v}$

Step 14 Given Information

(e)

The goal is to show that cross-product is not synonymous with associative.

The value of $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})$ is : $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v})\mathbf{w}\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Step 15 Explanation

 $(\mathbf{u}\times \mathbf{v})\times \mathbf{w}=(\mathbf{u}·\mathbf{v})\mathbf{w}-(\mathbf{u}·\mathbf{w})\mathbf{v}\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$ has the following value:

  From equation $\left(1\right)$ and $\left(2\right)$

 $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})\ne (\mathbf{u}\times \mathbf{v})\times \mathbf{w}$

  As a result, cross-product is not an associative term.

Step 16 Given Information

(f)

The goal is to determine whether $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$

Only when $(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v})\mathbf{w}=(\mathbf{u}·\mathbf{v})\mathbf{w}-(\mathbf{u}·\mathbf{w})\mathbf{v}$

$2\left[\right(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v}\left)\mathbf{w}\right]=0$

$(\mathbf{u}·\mathbf{w})\mathbf{v}-(\mathbf{u}·\mathbf{v})\mathbf{w}=0$
$(\mathbf{u}-\mathbf{w})\mathbf{v}=\mathbf{0}$ and $\mathbf{u}·\mathbf{v}=0$

 $\mathbf{u}·\mathbf{w}=0$ and $\mathbf{u}·\mathbf{v}=0$

The vector $\mathbf{u}$ must be orthogonal to both $\mathit{w}$ and $\mathit{v}$ in order for $\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}\times \mathbf{v})\times \mathbf{w}$