Question

The resultant of two vectors u and v is perpendicular to u. If |v|=2^1\2 |u| , show that the resultant of 2u and v is perpendicular to v.

Answer 1

Answer:

If the angle between vectors $\vec u$ and  $\vec u$ is $\theta$ then the angle made by the resultant with $\vec u$ is given by

$\boxed{\phi=\tan^{-1}\frac{v\sin\theta}{u+v\cos\theta}}$

Since the angle $\phi=90^\circ$

Therefore,

$u+v\cos\theta=0$

$\implies \cos\theta=-\frac{u}{v}$

But given that

$v=\sqrt{2}u$

Thus

$\cos\theta=-\frac{1}{\sqrt{2}}$

$\implies \cos\theta=\cos135^\circ$

$\implies \theta=135^\circ$

Now if the magnitude of the vector $\vec u$ is doubled then if the angle between $\vec v$ and the resultant is $\phi'$

Then

$\phi'=\tan^{-1}\frac{2u\sin135^\circ}{v+2u\cos135^\circ}$

$\implies \tan\phi'=\frac{2u\sin135^\circ}{v+2u\cos135^\circ}$

Now the value of

$v+2u\cos135^\circ$

$=v+2u\times (-\frac{1}{\sqrt{2}})$

$=v-\sqrt{2}u$

$=0$          (As given in the question)

$\implies \tan\phi'=\infty}$

or, $\phi'=90^\circ$

Therefore, the resultant of the vector $2\vec u$ and $\vec v$ is perpendicular to $\vec v$

Hope this helps.