​given that P+Q=R and R is perpendicular to P. If |P|=|R| , then what is the angle between P and Q?

1. pi/4
2. pi/2
3. 3pi/4
4. pi

Where P,Q and R are vectors.

Dear Student,

Please find below the solution to the asked query:

Let the angle between P & Q is θ. So,

$$\begin{gathered} R=\sqrt{P^2+Q^2+2PQ \cos \theta }\Rightarrow P^2=P^2+Q^2+2PQ cos \theta \\ \Rightarrow Q^2+2PQ cos \theta =0\Rightarrow Q+2P cos \theta =0 \\ \Rightarrow 2P cos \theta =-Q \end{gathered}$$

Given that R is perpendicular to P. Therefore,

$$\begin{gathered} \tan {90}^0=\frac{Q \sin \theta }{P+Q \cos \theta }\Rightarrow P+Q cos \theta =0 \\ \Rightarrow P+\left(-2P \cos \theta \right) cos \theta =0 \\ \Rightarrow P-2P co{s}^2 \theta =0 \\ \Rightarrow 2co{s}^2 \theta =1\Rightarrow \cos \theta =\frac{1}{\sqrt{2}} \\ \Rightarrow \theta ={45}^0=\frac{\pi }{4} \end{gathered}$$

Hope this information will clear your doubts about the topic.

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