Question

1. Two vectors of equal magnitude P are inclined at some angle such that the difference in magnitude
of resultant and magnitude of either of the vectors is 0.732 times either of the magnitude of vectors.
If the angle between them is increased by half of its initial value then find the magnitude of
difference of the vectors
1) 2 P
2) root2(P)
3) 3 P
4) root3P
~~~~~~~~~~~~vectors~~~~~~~~~~~~~​

Answer 1

Let R be the resultant vector.

Given that the difference in P and R is 0.732 times P.

$\sf{\longrightarrow R-P=0.732\,P}$

$\sf{\longrightarrow R=P+0.732\,P}$

$\sf{\longrightarrow R=1.732\,P}$

Taking $\sf{1.732=\sqrt3,}$

$\sf{\longrightarrow R=P\sqrt3\quad\quad\dots(1)}$

Let $\theta$ be the angle between the two vectors of magnitude P.

Then the magnitude of the resultant is given by,

$\sf{\longrightarrow R=\sqrt{P^2+P^2+2\cdot P\cdot P\cos\theta}}$

$\sf{\longrightarrow R=\sqrt{2P^2+2P^2\cos\theta}}$

$\sf{\longrightarrow R=\sqrt{2P^2(1+\cos\theta)}}$

$\sf{\longrightarrow R=\sqrt{2P^2\cdot2\cos^2\left(\dfrac{\theta}{2}\right)}}$

$\sf{\longrightarrow R=2P\cos\left(\dfrac{\theta}{2}\right)\quad\quad\dots(2)}$

From (1),

$\sf{\longrightarrow P\sqrt3=2P\cos\left(\dfrac{\theta}{2}\right)}$

$\sf{\longrightarrow \cos\left(\dfrac{\theta}{2}\right)=\dfrac{\sqrt3}{2}}$

$\sf{\longrightarrow\dfrac{\theta}{2}=30^o}$

$\sf{\longrightarrow\theta=60^o}$

Now the angle between the vectors of magnitude P each is increased by its half. So the new angle is,

$\sf{\longrightarrow\theta'=60^o+\dfrac{60^o}{2}}$

$\sf{\longrightarrow\theta'=90^o}$

So the angle between difference of the vectors becomes $\sf{90^o-180^o=-90^o.}$

Then the magnitude of difference of the vectors will be, from (2),

$\sf{\longrightarrow R'=2P\cos\left(\dfrac{-90^o}{2}\right)}$

$\sf{\longrightarrow\underline{\underline{R'=P\sqrt2}}}$

Answer 2

ANSWER : ROOT 2 P

The explanation is in the pics.