Question

If momentum of two particles of a system are given by:
${\overrightarrow{P}}_1=2\hat{i}-\hat{j}+3\hat{k}$ and ${\overrightarrow{P}}_2=-\hat{i}+2\hat{j}+3\hat{k}$, then the angle made by the direction of motion of the system with x-axis is :

• A. $co{s}^{-1}\frac{5}{14}$
• B. ${\cos }^{-1}\sqrt{36/38}$
• C. ${45}^{\circ }$
• D. ${\cos }^{-1}\sqrt{1/38}$

Answer 1

The correct option is A $co{s}^{-1}\frac{5}{14}$

We know dot product of two vectors are given as

$\overrightarrow{P_1}.\overrightarrow{P_2}=|\overrightarrow{P_1}||\overrightarrow{P_2}|\cos \theta$-------(A)

Given $\overrightarrow{P_1}=2\hat{i}-\hat{j}+3\hat{k}$

and $\overrightarrow{P_2}=-\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{P_1}.\overrightarrow{P_2}=5$

$|\overrightarrow{P_1}|.|\overrightarrow{P_2}|=\sqrt{2^2+\left({-1}^2\right)+3^2}.\sqrt{\left({-1}^2\right)+2^2+3^2}=14$

Putting the above values in A we get,

$\cos \theta =\frac{5}{14}$

$\theta ={\cos }^{-1}\left(\frac{5}{14}\right)$