Question

What is the angle between and the resultant of (P+)
and (P-O?
(a) Zero
(b) tan (PQ)
(c) tan-'(Q/P)
(d) tan (P-Q)/(P+Q)
please give explaination​

Answer 1

Correct Question:

What is the angle between $\sf{\vec{P}}$ and the resultant of $\sf{\left(\vec{P}+\vec{Q}\right)\:and\:\left(\vec{P}-\vec{Q}\right)}$

Answer:

Resultant of $\sf{\left(\vec{P}+\vec{Q}\right)\:and\:\left(\vec{P}-\vec{Q}\right)}$

$\sf{R=\vec{P}+\vec{Q}+\vec{P}-\vec{Q}=2\vec{P}}$

Now we have to find angle between

$\sf{2\vec{P}\:and\:\vec{P}}$

Resultant(R') of $\sf{2\vec{P}\:and\:\vec{P}=3\vec{P}}$

We know that,

$\sf{R^2=A^2+B^2+2ABcos\theta}$

Here $\theta$ is the angle to be found.

$\therefore\sf{(3P)^2=(2P)^2+(P)^2+2(2P)(P)cos\theta}$

$\sf{9P^2=4P^2+P^2+4P^2cos\theta}$

$\sf{9P^2-5P^2=4P^2cos\theta}$

$\sf{cos\theta=\dfrac{4P^2}{4P^2}}$

$\sf{cos\theta=1}$

$\sf{\theta=0^{\circ}}$

Therefore answer is (a) Zero