Question

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

• A. ${\cos }^{-1}\left(a\right)$
• B. ${\cos }^{-1}\left(b\right)$
• C. ${\cos }^{-1}\left(\frac{a}{\sqrt{a^2+b2}}\right)$
• D. ${\cos }^{-1}\left(\frac{b}{\sqrt{a^2+b2}}\right)$

Answer 1

The correct option is D

${\cos }^{-1}\left(\frac{b}{\sqrt{a^2+b2}}\right)$

Explanation for the correct answer:

Find the angle between $u$and $w$ :

Given,

$u=av+bw$

Multiply both sides with $w$,

Then,

$$\begin{gathered} u.w=av.w+bw.w \\ =a\left(0\right)+b{w}^2\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{v}\mathbf{,}\mathbf{w}\mathbf{}\mathbf{a}\mathbf{r}\mathbf{e}\mathbf{}\mathbf{\perp }\mathbf{]} \\ \Rightarrow u.w=b\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{w}\mathbf{.}\mathbf{w}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{]}\mathbf{}\mathbf{} \\ \Rightarrow \left|u\right|\left|w\right|\cos \theta =b \\ \Rightarrow \sqrt{(a^2+b^2)}\cos \theta =b\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{\mid }\mathbf{w}\mathbf{\mid }\mathbf{=}\mathbf{1}\mathbf{]} \\ \Rightarrow \cos \theta =\left(\frac{b}{\sqrt{a^2+b2}}\right) \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{b}{\sqrt{a^2+b2}}\right) \end{gathered}$$

Hence the correct answer is option D.