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 $\hat{w}\cdot (\hat{u}\times \vec{v})=1$. 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 $\vec{v}$
• C. If $\hat{u}$ lies in the $xy$-plane then $\left|u_1\right|=\left|u_2\right|$
• D. If $\hat{u}$ lies in the $xz$-plane then $2\left|u_1\right|=\left|u_3\right|$

Answer 1

Answer: (B), (C)

Let $\vec{v}=x\hat{i}+y\hat{j}+z\hat{k}$
$\hat{w}\cdot (\vec{u}\times \vec{v})=1$
$|\hat{w}\Vert \vec{u}\times \vec{v}|\cos \theta =1$
$\Rightarrow \cos \theta =1$
$\theta =0$
$\therefore \hat{w}\Vert \vec{u}\times \vec{v}$
$\hat{w}\cdot \vec{u}=0$
$\Rightarrow u_1+u_2+2u_3=0$
$\hat{w}\cdot \vec{v}=0$
$\Rightarrow x+y+2z=0$
So there are infinite many choices for such $\vec{v}$
If $\hat{u}$ lies in $xy$ plane then
$u_1+u_2=0$
$u_2=-u_1$
$\Rightarrow \left|u_2\right|=\left|u_1\right|$
If $\hat{u}$ lies in $xz$ plane
$u_1+2u_3=0$
$\left|u_1\right|=2\left|u_3\right|$