Question

$\overrightarrow{a}=6\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b}=2\hat{i}-9\hat{j}+6\hat{k}$ then find $\overrightarrow{a}\cdot \overrightarrow{b}$ and the angle between $a$ and $b$

Answer 1

$\overrightarrow{a}\cdot \overrightarrow{b}=(6\hat{i}+2\hat{j}+3\hat{k})\cdot (2\hat{i}-9\hat{j}+6\hat{k})$

$=12-18+18$ since $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$

$=12$

$\cos \theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{12}{\sqrt{36+4+9}\sqrt{4+81+36}}$

$=\frac{12}{7\times 11}=\frac{12}{77}$

$\theta ={\cos }^{-1}\left(\frac{12}{77}\right)$.