Question

If $\overline{v}$ and $\overline{w}$ are two mutually perpendicular unit vectors and $\overline{u}=a\overline{v}+b\overline{w}$, where a and b are non zero real numbers, then the angle between $\overline{u}$ and $\overline{w}$ 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

Answer: (A)

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

We have,

$\left[\overline{v}\right]=\left[\overline{u}\right]=1\overline{u}=a\overline{v}+b\overline{w}\left|u\right|=\sqrt{a^2+b^2}u+\theta betheanglebetween\overline{u}and\overline{w}\overline{u}.\overline{w}=\left|\overline{u}\right|\left|\overline{w}\right|\cos \theta \left(a\overline{v}+b\overline{w}\right).\overline{w}=\sqrt{a^2+b^2}\cos \theta b=\sqrt{a^2+b^2}\cos \theta \Rightarrow \cos \theta =\frac{b}{\sqrt{a^2+b^2}}\therefore \theta ={\cos }^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)$

So, correct option is A.