Question

If $| a |=1,| b |=4, a \cdot b =2$ and $c =2 a \times b -3 b$, then the angle between $b$ and $c$ is

• A. $\frac{\pi}{6}$
• B. $\frac{5 \pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{2 \pi}{3}$

Answer 1

Answer: (B)

$| a \times b |^{2}=| a |^{2}| b |^{2}-2( a \cdot b )^{2}=16-4=12$ and
$| c |^{2}=(2 a \times b -3 b )^{2}=(2 a \times b -3 b )$
$=4| a \times b |^{2}+9| b |^{2}=4.12+9.16$
$=192 \Rightarrow | c |=8 \sqrt{3}$
Now, $b \cdot c = b (2 a \times b -3 b )=-3| b |^{2}=-48$
$\therefore \cos \theta=\frac{ b \cdot c }{| b || c |}=-\frac{48}{4 \cdot 8 \sqrt{3}}=-\frac{\sqrt{3}}{2}$
$\Rightarrow \theta =\frac{5 \pi}{6}$