Question

If $3p+2q=i+j+k$ and $3p-2q=i-j-k$ then the angle between $p$ and $q$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$
• E. $\pi$

Answer 1

Answer: (D)

Given, $3p+2q=i+j+k\dots (i)$
and $3p-2q=i-j-k\dots (ii)$
On adding Eqs. (i) and (ii), we get
$6p=2i$
$\therefore p=\frac{i}{3}$
On subtracting Eq. (ii) from Eq. (i), we get
$4q=2(j+k)$
$\Rightarrow q=\frac{1}{2}(j+k)$
Let $\theta$ be the angle between $p$ and $q$, then
$p\cdot q=|p||q|\cos \theta$
$\Rightarrow \cos \theta =\frac{p\cdot q}{|p||q|}$
$\Rightarrow \cos \theta =\frac{1}{6}\frac{i\cdot (j+k)}{|p||q|}=0$
$=\cos 90^{\circ }$
$\therefore \theta =90^{\circ }$
$=\frac{\pi }{2}$