# vectors and scalars… [closed]

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

## closed as off-topic by Michael Rybkin, Paul Frost, Javi, Shailesh, José Carlos SantosApr 16 at 12:12

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• Welcome to Mathematics StackExchange! Have you tried to solve the problem? I would start by evaluating/simplifying the expression for $\hat{u}$. – Matti P. Apr 16 at 7:16
• HI, so I evaluated the problem by applying the triple multiplication law in cross product . i arrived at $$(\overrightarrow{a}.\hat{i})i-(\hat{i}.\hat{i})a+(\overrightarrow{a}.\hat{j})j-(\hat{j}.\hat{j})a+(\overrightarrow{a}.\hat{k})k-(\hat{k}.\hat{k})a$$ – gucci Apr 16 at 7:23
• @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. – Matti P. Apr 16 at 7:33
• yeah I did that too I got : $$(\overrightarrow{a}.\hat{i})i-a+(\overrightarrow{a}.\hat{j})j-a+(\overrightarrow{a}.\hat{k})k-a$$ – gucci Apr 16 at 7:38
• after this step i get lost – gucci Apr 16 at 7:38

## 1 Answer

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

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

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