The sum of two forces $$\vec{P}$$ and $$\vec{Q}$$ is $$\vec{R}$$ such that $$|\vec{R}| = |\vec{P}|$$. Find the angle between resultant of $$2\vec{P}$$ and $$\vec{Q}$$, and $$\vec{Q}$$,________

Let the magnitudes of the two given forces be denoted by $$|\vec P| = P$$ and $$|\vec Q| = Q$$.

Their resultant is $$\vec R = \vec P + \vec Q$$ and we are told that its magnitude equals that of $$\vec P$$, that is

$$|\vec R| = |\vec P|\; \Longrightarrow\; |\vec R| = P.$$

First we express the magnitude of the resultant $$\vec R$$ in terms of $$P$$, $$Q$$ and the angle $$\alpha$$ between $$\vec P$$ and $$\vec Q$$. The law of vector addition (or the triangle law) gives

$$|\vec R|^2 \;=\; |\vec P|^2 + |\vec Q|^2 + 2\,|\vec P|\,|\vec Q|\,\cos\alpha.$$

Substituting $$|\vec R| = P$$ and $$|\vec P|=P,\; |\vec Q|=Q$$ we have

$$P^2 = P^2 + Q^2 + 2PQ\cos\alpha.$$

Subtracting $$P^2$$ from both sides yields

$$0 = Q^2 + 2PQ\cos\alpha.$$

Dividing by the non-zero quantity $$Q$$ we get

$$Q + 2P\cos\alpha = 0.$$

Hence

$$\cos\alpha = -\dfrac{Q}{2P}.$$

Now we form a new resultant by doubling $$\vec P$$ and adding $$\vec Q$$. Let

$$\vec S = 2\vec P + \vec Q.$$

We wish to find the angle $$\theta$$ between $$\vec S$$ and $$\vec Q$$. By the definition of the dot product,

$$\vec S\cdot\vec Q = |\vec S|\,|\vec Q|\,\cos\theta.$$

Compute the dot product on the left:

$$\vec S\cdot\vec Q = (2\vec P + \vec Q)\cdot\vec Q = 2\,\vec P\cdot\vec Q + \vec Q\cdot\vec Q.$$

The individual dot products are

$$\vec P\cdot\vec Q = |\vec P|\,|\vec Q|\,\cos\alpha = P Q \cos\alpha,$$

$$\vec Q\cdot\vec Q = |\vec Q|^2 = Q^2.$$

Therefore

$$\vec S\cdot\vec Q = 2(P Q \cos\alpha) + Q^2.$$

Substituting the previously obtained value $$\cos\alpha = -\dfrac{Q}{2P}$$, we find

$$\vec S\cdot\vec Q = 2\!\left(P Q \left(-\dfrac{Q}{2P}\right)\right) + Q^2 = 2\!\left(-\dfrac{Q^2}{\,2\,}\right) + Q^2 = -Q^2 + Q^2 = 0.$$

Because $$\vec S\cdot\vec Q = 0$$, the cosine of the angle $$\theta$$ is

$$\cos\theta = \dfrac{\vec S\cdot\vec Q}{|\vec S|\,|\vec Q|} = \dfrac{0}{|\vec S|\,|\vec Q|} = 0.$$

An angle whose cosine is zero is a right angle, namely $$\theta = 90^\circ$$.

Hence, the correct answer is Option D (90°).