Question

Find the angle between $\overrightarrow{A}=4\hat{i}+\hat{j}+3\hat{k}$ and $\overrightarrow{B}=\hat{i}+3\hat{j}+4\hat{k}$

Answer 1

Given that,

$\overrightarrow{A}=4\hat{i}+\hat{j}+3\hat{k}$

$\overrightarrow{B}=\hat{i}+3\hat{j}+4\hat{k}$

We know that,

$\overrightarrow{A}\bullet \overrightarrow{B}=AB\cos \theta$

$cos\theta =\frac{\overrightarrow{A}\bullet \overrightarrow{B}}{AB}$

Now, put the value of A and B

$\overrightarrow{A}\bullet \overrightarrow{B}=4\hat{i}+\hat{j}+3\hat{k}\bullet \hat{i}+3\hat{j}+4\hat{k}$

$\overrightarrow{A}\bullet \overrightarrow{B}=4+3+12$

$\overrightarrow{A}\bullet \overrightarrow{B}=19$

Now,

$A=\sqrt{{\left(4\right)}^2+{\left(1\right)}^2+{\left(3\right)}^2}$

$A=\sqrt{26}$

$B=\sqrt{{\left(1\right)}^2+{\left(3\right)}^2+{\left(4\right)}^2}$

$B=\sqrt{26}$

Now,

$\cos \theta =\frac{19}{\sqrt{26}\times \sqrt{26}}$

$\cos \theta =\frac{19}{26}$

$\theta ={\cos }^{-1}\frac{19}{26}$

Hence the angle is ${\cos }^{-1}\frac{19}{26}$