Question

Find an expression for the sine of the angle between the two vectors $3\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k}$ and $2\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k}$.

Answer 1

Sine angle between vector $(3i+j+2k)$ & $(2i-2j+4k)$

$\to$ angle between two vector $\overrightarrow{a}$ & $\overrightarrow{b}$ is given by

$cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}$

$\to \overrightarrow{a}.\overrightarrow{b}=(3,1,2).(2,-2,4)$

$=6-2+8$

$=4+8$

$=12$

$\to \left|\overrightarrow{a}\right|=\sqrt{(3{)}^2+(1{)}^2+(2{)}^2}=\sqrt{9+1+4}=\sqrt{14}$

$\left|\overrightarrow{b}\right|=\sqrt{(2{)}^2+(-1{)}^2+(6{)}^2}=\sqrt{4+4+16}=\sqrt{24}$

$\to cos\theta =\frac{12}{\sqrt{14}.\sqrt{24}}$

$=\frac{12}{\sqrt{14}\sqrt{12\times 2}}$

$=\frac{\sqrt{12}.\sqrt{12}}{\sqrt{14}\sqrt{12}.\sqrt{2}}$

$=\frac{\sqrt{12}}{\sqrt{28}}$

$=\sqrt{\frac{12}{28}}$

$\to cos\theta =\sqrt{\frac{3}{7}}$

$\to si{n}^2\theta +co{s}^2\theta =1$

$\therefore si{n}^2\theta +\frac{3}{7}=1$

$\therefore si{n}^2\theta =1-\frac{3}{7}=\frac{7-3}{7}=\frac{4}{7}$

$\therefore sin\theta =\sqrt{\frac{4}{7}}$