Question

Let $u$ and $v$ be two vectors in $R^2$ whose Euclidean norms satisfy $\parallel U\parallel$=$2\parallel V\parallel$. What is the value of $\alpha$ such that $w=u+\alpha v$ bisects the angle between $u$ and $v$?

• A. 2
• B. 1/2
• C. 1
• D. -1/2

Answer 1

The correct option is A 2

$\because \parallel u\parallel =2\parallel v\parallel$ then we can ssume $(u=\frac{2}{0})$ and $v=\left(\frac{0}{1}\right)$ and let $w=\left(\frac{2}{\alpha }\right)$then $u.w=4$ and $v.w=\alpha$
$\Rightarrow |u||w|cos{\theta }_1=4$ and
$|v||w|cos{\theta }_2=\alpha$
$cos{\theta }_2=\frac{4}{2\sqrt{4+{\alpha }^2}}$
and $cos{\theta }_2=\frac{\alpha }{1.\sqrt{4+{\alpha }^2}}$
$\therefore w$ is the angle bisector between $u$ and $v$
So, $cos{\theta }_1=cos{\theta }_2$
$\Rightarrow \frac{4}{2\sqrt{4+{\alpha }^2}}=\frac{\alpha }{1.\sqrt{4+{\alpha }^2}}$
$\alpha =2$