Question

If $|a|=1,|b|=4,a\cdot b=2$ and $c=2a\times b-3b$, 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
$|c{|}^2=(2a\times b-3b{)}^2=(2a\times b-3b)$
$=4|a\times b{|}^2+9|b{|}^2=4.12+9.16$
$=192\Rightarrow |c|=8\sqrt{3}$
Now, $b\cdot c=b(2a\times b-3b)=-3|b{|}^2=-48$
$\therefore \cos \theta =\frac{b\cdot c}{|b||c|}=-\frac{48}{4\cdot 8\sqrt{3}}=-\frac{\sqrt{3}}{2}$
$\Rightarrow \theta =\frac{5\pi }{6}$