Question

Find the difference in the value of the vector, if the two with equal magnitude $\overrightarrow{P}$ are inclined at an angle such that the difference in the magnitude of the resultant and magnitude of either vector is $0.732$ times either vector. Also, the angle between them is increased to half of its initial.

Answer 1

Step 1: Given

Given that the two vectors have equal magnitude $P$.

Let us assume that the angle between them is $\theta$.

If $\overrightarrow{R}$ is the resultant vector, then it is given that

$\left|\overrightarrow{R}\right|-P=0.732P$

Step 2: Calculate the Resultant Vector

The magnitude of the resultant vector $\overrightarrow{R}$ can be given as,

$\left|\overrightarrow{R}\right|=\sqrt{P^2+P^2+2P^2\cos \theta }$ $\left[\because \left|\overrightarrow{R}\right|=\sqrt{A^2+B^2+2AB\cos \theta }\right]$

$=P\sqrt{2(1+\cos \theta )}$

$=P\sqrt{4{\cos }^2\frac{\theta }{2}}$ $\left[\because 1+\cos \theta =2{\cos }^2\frac{\theta }{2}\right]$

$=2P\cos \left(\frac{\theta }{2}\right)$

Step 3: Calculate the Angle $\theta$

Substitute the value of the resultant vector in the established equation to get,

$2P\cos \frac{\theta }{2}-P=0.732P$

$\Rightarrow 2\cos \frac{\theta }{2}-1=0.732$

$\Rightarrow 2\cos \frac{\theta }{2}=1+0.732=1.732$

$\Rightarrow \cos \frac{\theta }{2}=\frac{\sqrt{3}}{2}$ $\left[\because \sqrt{3}=1.732\right]$

$\Rightarrow$ $\frac{\theta }{2}=30°$ $\left[\because \cos 30°=\frac{\sqrt{3}}{2}\right]$

$\Rightarrow$ $\theta =60°$

Step 4: Calculate the Difference Vector

Given that the angle between them is increased to half of its initial.

$\therefore \theta \text{'}=60+\frac{60}{2}=90°$

Hence, the difference between the vectors can be given as,

$\left|\overrightarrow{R\text{'}}\right|=\sqrt{P^2+P^2-2P^2\cos 90°}$ $\left[\because \left|\overrightarrow{R\text{'}}\right|=\sqrt{A^2+B^2-2AB\cos \theta }\right]$

$=\sqrt{2}P$ $\left[\because \cos 90°=0\right]$

Hence, the required difference in the value of the vector is $\sqrt{2}P$.