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A vector $\mathbf{P}$ directed along the $X$-axis is added to vector $\mathbf{Q}$ which has a magnitude of $10 \mathrm{~m}$. The resultant vector is directed along the $Y$-axis, with a magnitude that is 2 times that of $\mathbf{P}$. The magnitude of $\mathbf{P}$ is
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The correct answer is:
$2 \sqrt{5} \mathrm{~m}$
By parallelogram law of vector addition, angle made by resultant with $\mathbf{P}$ is given by;
$\tan \alpha=\frac{Q \sin \theta}{P+Q \cos \theta}$
Here $\alpha=90^{\circ}$ so ; $\tan 90^{\circ}=\infty=\frac{Q \sin \theta}{P+Q \cos \theta}$
$\Rightarrow \quad P+Q \cos \theta=0$ or $Q \cos \theta=-P$
Now, magnitude of resultant,
$R=\sqrt{P^2+Q^2+2 P Q \cos \theta}$
Here, we have $R=2 P$ and $Q \cos \theta=-P$ and $Q=10 \mathrm{~m}$
So, $\quad 2 P=\sqrt{P^2+100-2 P^2}$
$\Rightarrow \quad 4 P^2=100-P^2 \Rightarrow 5 P^2=100$
or $\quad P=\sqrt{20}=2 \sqrt{5} \mathrm{~m}$
Magnitude of $P$ is $2 \sqrt{5} \mathrm{~m}$
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