Question

Calculate the scalar product of the following vectors.
Find the angle between the vectors $2aandb/2ifa=\{-4,2,4\}andb=\{\sqrt{2},\sqrt{-2},0\}$

Answer 1

Given $a=(-4,2,4);b=(\sqrt{2},-\sqrt{2},0)$

$2a.\frac{b}{2}=a.b=|a||b|\cos \theta$

$-4\sqrt{2}-2\sqrt{2}+0=|\sqrt{4^2+2^2+4^4}||\sqrt{(\sqrt{2}{)}^2+(\sqrt{2}{)}^2+0}|\cos \theta$

$-6\sqrt{2}=12\cos \theta$

$\cos \theta =\frac{-1}{\sqrt{2}}$

$\theta=\dfrac{3\pi}{4}=\dfrac{\pi}{4}$