Question

Let $\hat{u}=u_1\hat{i}+u_2\hat{j}+u_3\hat{k}$ be a unit vector in $R^3$ and $\hat{w}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k}).$ Given that there exists a vector $\overrightarrow{v}$ in $R^3$ such that $|\hat{u}\times \overrightarrow{v}|=1$ and $\hat{w}\cdot (\hat{u}\times \overrightarrow{v})=1.$ Which of the following statement(s) is(are) correct ?

• A. There is exactly one choice for such $\overrightarrow{v}$
• B. There are infinitely many choices for such $\overrightarrow{v}$
• C. If $\hat{u}$ lies in the $xy$-plane then $|u_1|=|u_2|$
• D. If $\hat{u}$ lies in the $xz$-plane then $2|u_1|=|u_3|$

Answer 1

Answer: (B), (C)

The correct options are
B There are infinitely many choices for such $\overrightarrow{v}$
C If $\hat{u}$ lies in the $xy$-plane then $|u_1|=|u_2|$

$\hat{u}=u_1\hat{i}+u_2\hat{j}+u_3\hat{k}$
$\hat{w}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})$
$|\hat{u}\times \overrightarrow{v}|=1$

$\hat{w}\cdot (\hat{u}\times \overrightarrow{v})=1$
$\Rightarrow |\hat{w}||\hat{u}\times \overrightarrow{v}|\cos \alpha =1\Rightarrow \cos \alpha =1\Rightarrow \alpha =0^{\circ }$
So, $\hat{w}\perp \hat{u}$ and $\hat{w}\perp \overrightarrow{v}\cdots \left(i\right)$
$\therefore \hat{u}$ and $\overrightarrow{v}$ lies in the same plane.
So there will be infinite number of such $\overrightarrow{v}.$

Option (3):
If $\hat{u}$ lies in the $xy$-plane,
$\hat{u}=u_1\hat{i}+u_2\hat{j}\hat{u}\cdot \hat{w}=0\left[\text{From}\right(i\left)\right]\Rightarrow u_1+u_2=0\Rightarrow |u_1|=|u_2|$

Option (4):
If $\hat{u}$ lies in the $xz$-plane,
$\hat{u}=u_1\hat{i}+u_3\hat{k}$
$\hat{u}\cdot \hat{w}=0\Rightarrow u_1+2u_3=0\Rightarrow 2|u_3|=|u_1|$