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
$cos^{2} \theta+cos^{2} \theta+p^{2} =1$
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 $\theta$ )
$\therefore \frac{\pi}{4} \le cos \,\theta\le \frac{3\pi}{4}$