Let $\vec u= u_1\hat i+u_3\hat k$ be a unit vector in xz-plane and $\vec w =\frac{1}{\sqrt{6}} (\hat i+\hat j+2\hat k)$. If there exists a vector $\vec v$ in such that $|\vec u×\vec v|=1$ and $\vec w. (\vec u×\vec v)$. Then,
Answer & explanation
Correct answer: option 2
We observe that $\vec w$ is a unit vector.
Now, $|\vec u×\vec v|=1$ and $\vec w. (\vec u×\vec v)$
$⇒\vec w=\vec u×\vec v$
$⇒\vec w=\begin{vmatrix}\hat i&\hat j&\hat k\\u_1&0&u_3\\v_1&v_2&v_3\end{vmatrix}$
$⇒\frac{1}{\sqrt{6}}(\hat i+\hat j+2\hat k)=-u_3\,v_2\hat i-(u_1\,v_3-u_3\,v_1)\hat j+u_1\,v_2\hat k$
$⇒u_3\,v_2=-\frac{1}{\sqrt{6}},-(u_1\,v_3-u_3\,v_1)=\frac{1}{\sqrt{6}}$ and $u_1\,v_2=\frac{2}{\sqrt{6}}$
$⇒u_1\,v_2=-2u_3\,v_2$
$⇒u_1=-2u_3⇒|u_1|=2|u_3|$