Question

Let $\vec{a}.\vec{b}=0,$ where $\vec{a}$ and $\vec{d}$ are unit vectors and the unit vector $\vec{c}$ is inclined at an angle $\theta$ to both $\vec{a}$ and $\vec{d}$
If $\vec{c}=m\vec{a}+n\vec{b}+P\left(\vec{a}\times \vec{b}\right),\left(m,n,p\in R\right),$then

• A. $-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}$
• B. $\frac{\pi }{4}\le \theta \le \frac{3\pi }{4}$
• C. $0\le \theta \le \frac{\pi }{4}$
• D. $0\le \theta \le \frac{3\pi }{4}$

Answer 1

Answer: (B)

$\vec{c}=m\vec{a}+n\vec{b}+p\left(\vec{a}\times \vec{b}\right)$
Taking dot product with $\vec{a}$ and $\vec{b}$ we have
$m=n=cos\theta$
$\Rightarrow \left|\vec{c}\right|=\left|cos\theta \vec{a}+cos\theta \vec{b}+p\left(\vec{a}\times \vec{b}\right)\right|=1$
Squaring both sides, we get
$co{s}^2\theta +co{s}^2\theta +p^2=1$
or $\theta =\pm \frac{\sqrt{1-p^2}}{\sqrt{2}}$
Now $-\frac{1}{\sqrt{2}}\le cos\theta \le \frac{1}{\sqrt{2}}$ (for real value of $\theta$ )
$\therefore \frac{\pi }{4}\le cos\theta \le \frac{3\pi }{4}$