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 $x y$-plane then $\left| u _{1}\right|=\left| u _{2}\right|$
• D. If $\hat{u}$ lies in the $x z$-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} \| \vec{u} \times \vec{v}| \cos \theta=1 $
$\Rightarrow \cos \theta=1 $
$ \theta=0 $
$\therefore \hat{w} \| \vec{u} \times \vec{v} $
$\hat{w} \cdot \vec{u}=0 $
$\Rightarrow u_{1}+u_{2}+2 u_{3}=0 $
$\hat{w} \cdot \vec{v}=0 $
$\Rightarrow x+y+2 z=0$
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 $x z$ plane
$ u _{1}+2 u_{3}=0$
$\left| u _{1}\right|=2\left| u _{3}\right|$