Question

If $p$ and $q$ are non-collinear unit vectors and $|p+q|=\sqrt{3}$, then $(2p-3q)\cdot (3p+q)$ is equal to

• A. $0$
• B. $\frac{1}{3}$
• C. $-\frac{1}{3}$
• D. $\frac{1}{2}$
• E. $-\frac{1}{2}$

Answer 1

Answer: (E)

$-\frac{1}{2}$

Given, $|p+q|=\sqrt{3}\Rightarrow |p+q{|}^2=(\sqrt{3}{)}^2$
$\Rightarrow |p{|}^2+|q{|}^2=2p\cdot q=3$
$(\because p$ and $q$ are non-collinear unit vectors, i.e., $|p|=|q|=1)$
$\Rightarrow 1+1+2p\cdot q=3$
$\Rightarrow p\cdot q=\frac{1}{2}$ .... (i)
Now $(2p-3q)\cdot (3p+q)$
$=6|p{|}^2--9p\cdot q+2p\cdot q-3|q{|}^2$
$=6\left(1\right)-7p\cdot q-3\left(1\right)$
$=3-7\left(\frac{1}{2}\right)$ ..... [from Eq. (i)]
$=-\frac{1}{2}$