Discussion Forum

If $\vec{a}=(\hat{i}+\hat{j}+\hat{k})$, $\vec{a}.\vec{b} = 1$ and $\vec{a}\times\vec{b} = \hat{j}-\hat{k}$,

then $\vec{b}$ is

Answer:

Explanation:

$\because(\vec{a}\times\vec{b})\times\vec{a}=(\vec{a}.\vec{a})\vec{b}-(\vec{a}.\vec{b})\vec{a}$

$\therefore(\hat{j}-\hat{k})\times(\hat{i}+\hat{j}+k)=(\sqrt{3})^{2}(\vec{b})-(\hat{i}+\hat{j}+k)$

$\Rightarrow 3\hat{b}=3\hat{i}\Rightarrow \hat{b}=\hat{i}$