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If $\vec u=\hat i×(\vec a×\hat i)+\hat j×(\vec a×\hat j)+\hat k×(\vec a×\hat k)$, then:
(A) $\vec u$ is a unit vector
(B) $\vec u=\vec a+\hat i+\hat j+\hat k$
(C) $\vec u=2\vec a$
(D) $\vec u=8(\hat i+\hat j+\hat k)$

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  • $\begingroup$ Welcome to Mathematics StackExchange! Have you tried to solve the problem? I would start by evaluating/simplifying the expression for $\hat{u}$. $\endgroup$ – Matti P. Apr 16 at 7:16
  • $\begingroup$ HI, so I evaluated the problem by applying the triple multiplication law in cross product . i arrived at $$ (\overrightarrow{a}.\hat{i})i-(\hat{i}.\hat{i})a+(\overrightarrow{a}.\hat{j})j-(\hat{j}.\hat{j})a+(\overrightarrow{a}.\hat{k})k-(\hat{k}.\hat{k})a $$ $\endgroup$ – gucci Apr 16 at 7:23
  • $\begingroup$ @gucci Good! Can you evaluate it further? For example, $\hat{i} \cdot \hat{i} = ??$ ... I think here we can assume that these are the base vectors of the coordinate system. $\endgroup$ – Matti P. Apr 16 at 7:33
  • $\begingroup$ yeah I did that too I got : $$ (\overrightarrow{a}.\hat{i})i-a+(\overrightarrow{a}.\hat{j})j-a+(\overrightarrow{a}.\hat{k})k-a $$ $\endgroup$ – gucci Apr 16 at 7:38
  • $\begingroup$ after this step i get lost $\endgroup$ – gucci Apr 16 at 7:38
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$\vec u=\hat i×(\vec a×\hat i)+\hat j×(\vec a×\hat j)+\hat k×(\vec a×\hat k)$

We have $$\hat i×(\vec a×\hat i)=(\hat i . \hat i)\vec a- (\hat i. \vec a)\hat i= \vec a -(\hat i. \vec a)\hat i$$

And similarly, $$\hat j×(\vec a×\hat j)=\vec a- (\hat j. \vec a)\hat j$$ $$\hat k ×(\vec a×\hat j)=\vec a- (\hat k. \vec a)\hat k$$

But, $(\hat i. \vec a)$ is the abscissa of $\vec a$ (i.e. the first component of the vector $\vec a$ i.e. $a_x$)

So, $(\hat i. \vec a)\hat i +(\hat j. \vec a)\hat j+ (\hat k. \vec a)\hat k=\vec {a_x}+ \vec {a_y}+ \vec {a_z} =\vec a$

Thus $$\vec u=\vec a + \vec a +\vec a -\vec a= 2\vec a$$

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  • $\begingroup$ thanks a lot..... $\endgroup$ – gucci Apr 16 at 8:28
  • $\begingroup$ Your welcome :-) $\endgroup$ – Fareed AF Apr 16 at 9:16

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