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# Two vectors P and Q have equal magnitudes.

Question:

Two vectors $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ have equal magnitudes. If the magnitude of $\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}$ is $n$ times the magnitude of $\vec{P}-\vec{Q}$, then angle between $\vec{P}$ and $\vec{Q}$ is :

1. $\sin ^{-1}\left(\frac{\mathrm{n}-1}{\mathrm{n}+1}\right)$

2. $\cos ^{-1}\left(\frac{\mathrm{n}-1}{\mathrm{n}+1}\right)$

3. $\sin ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)$

4. $\cos ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)$

Correct Option: , 4

Solution:

$|\overrightarrow{\mathrm{P}}|=|\overrightarrow{\mathrm{Q}}|=\mathrm{x}$ .......$\ldots$ (i)

$|\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}|=\mathrm{n}|\overrightarrow{\mathrm{P}}-\overrightarrow{\mathrm{Q}}|$

$\mathbf{P}^{2}+\mathbf{Q}^{2}+2 \mathrm{PQ} \cos \theta=\mathbf{n}^{2}\left(\mathrm{P}^{2}+\mathbf{Q}^{2}-2 \mathrm{PQ} \cos \theta\right)$

Using (i) in above equation

$\cos \theta=\frac{n^{2}-1}{1+n^{2}}$

$\theta=\cos ^{-1}\left(\frac{n^{2}-1}{n^{2}+1}\right)$