Question

given vectors A=-i+2j-3k and b=4i-2j+6k the angle made (A+B) with axis is​

Answer 1

Answer:

$\bf{\pi/4}$

Explanation:

Given:

$\vec{a}=-\vec{i}+2\vec{j}-3\vec{k}$

$\vec{b}=4\vec{i}-2\vec{j}+6\vec{k}$

$\vec{a}+\vec{b}=3\vec{i}+0\vec{j}+3\vec{k}$

$\text{Let}\:\theta\:\text{be the angle made by}\:\vec{a}+\vec{b}\:\text{with the x axis}$

$cos\theta=\frac{(\vec{a}+\vec{b}).\vec{i}}{|\vec{a}+\vec{b})|\:|\vec{i}|}$

$cos\theta=\frac{3}{(\sqrt{9+9})\:(1)}$

$cos\theta=\frac{3}{3\sqrt{2}}$

$cos\theta=\frac{1}{\sqrt2}$

$\bf{\theta=\pi/4}$