vectors and scalars....

Question

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) $\vec u=8(\hat i+\hat j+\hat k)$

Answer 1

$\vec u=\hat i×(\vec a×\hat i)+\hat j×(\vec a×\hat j)+\hat k×(\vec a×\hat k)$

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