Question

If $v$ and $w$ are two mutually perpendicular unit vectors and $u=av+bw$, where $a$ and $b$ are non-zero real number, then the angle between $u$ and $w$ is

• A. ${\cos }^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)$
• B. ${\cos }^{-1}\left(\frac{a}{\sqrt{a^2+b^2}}\right)$
• C. ${\cos }^{-1}\left(b\right)$
• D. ${\cos }^{-1}\left(a\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)$

We have, $u=av+bw$

$u.w=av.w+bw.w=a\left(0\right)+b{w}^2$ $(\because v,w$ are $\perp )$

$=b$ $(\because w$ is unit vecotr $)$

$\Rightarrow \left|u\right|\left|w\right|\cos \theta =b\Rightarrow \sqrt{a^2+b^2}\cos \theta =b(\because \left|w\right|=1)$

$\Rightarrow \cos \theta =\frac{b}{\sqrt{a^2+b^2}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)$