Question

The resultant ${\displaystyle \overrightarrow{\text{R}}}$ of ${\displaystyle \overrightarrow{\text{P}}}$ and ${\displaystyle \overrightarrow{\text{Q}}}$ is perpendicular to ${\displaystyle \overrightarrow{\text{P}}}$. Also ${\displaystyle \left|\overrightarrow{\text{P}}\right|=\left|\overrightarrow{\text{R}}\right|}$. The angle between ${\displaystyle \overrightarrow{\text{P}}}$ and ${\displaystyle \overrightarrow{\text{Q}}}$ is ______.
[tan 45° = 1]

पर्याय

• ${\displaystyle \frac{5\pi }{4}}$

• ${\displaystyle \frac{7\pi }{4}}$

• ${\displaystyle \frac{\pi }{4}}$

• ${\displaystyle \frac{3\pi }{4}}$

Answer 1

The resultant ${\displaystyle \overrightarrow{\text{R}}}$ of ${\displaystyle \overrightarrow{\text{P}}}$ and ${\displaystyle \overrightarrow{\text{Q}}}$ is perpendicular to ${\displaystyle \overrightarrow{\text{P}}}$. Also ${\displaystyle \left|\overrightarrow{\text{P}}\right|=\left|\overrightarrow{\text{R}}\right|}$. The angle between ${\displaystyle \overrightarrow{\text{P}}}$ and ${\displaystyle \overrightarrow{\text{Q}}}$ is ${\displaystyle \underline{\frac{3\pi }{4}}}$.

Explanation:

Given that R is resultant of ${\displaystyle \overrightarrow{\text{P}}}$ and ${\displaystyle \overrightarrow{\text{Q}}}$ as shown in the figure below,

BC = P, CA = Q and BA = R

Given, BA and BC are perpendicular and equal in magnitude. So, from property of triangle,

∠ACB = 45°

Now, BC has to be extended up to D so,

that CD = P

Now, CD and CA have the initial point C, so the angle between CD and CA;

= 180° - 45° = 135° = ${\displaystyle \frac{3\pi }{4}}$

So, angle between ${\displaystyle \overrightarrow{\text{P}}}$ and ${\displaystyle \overrightarrow{\text{Q}}}$ is ${\displaystyle \frac{3\pi }{4}}$.