Question

If $\overrightarrow{A}=2\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{B}=\sqrt{2}(\hat{i}+\hat{j})$. Find $\overrightarrow{A}.\overrightarrow{B}$. Hence find the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$. Also find the component of $\overrightarrow{A}$ along $\overrightarrow{B}.$

Answer 1

Given that,

$\overrightarrow{A}=2\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{B}=\sqrt{2}(\hat{i}+\hat{j})$

We know that,

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

$\cos \theta =\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}....\left(I\right)$

Now, the dot product of $\overrightarrow{A}$ and $\overrightarrow{B}$ is

$\overrightarrow{A}\cdot \overrightarrow{B}=(2\hat{i}+\hat{j}-\hat{k})\cdot (\sqrt{2}\hat{i}+\sqrt{2}\hat{j})$

$\overrightarrow{A}\cdot \overrightarrow{B}=2\sqrt{2}+\sqrt{2}$

Now, $A$ and $B$ are the magnitude of the vecotor $\overrightarrow{A}$ and $\overrightarrow{B}$

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

$A=\sqrt{6}$

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

$B=\sqrt{4}=2$

Now, put the value in equation (I)

$\cos \theta =\frac{2\sqrt{2}+\sqrt{2}}{\sqrt{6}\sqrt{4}}$

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

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

$\theta ={30}^0$

Hence, the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is ${30}^0$