Numerical Problems Based on Composition of Vectors for Class 11 Physics

Let the angle between vector $\vec{P}$ and vector $\vec{Q}$ be $\theta$.

The magnitude of the resultant vector $\vec{R}$ is given by the parallelogram law of vector addition:

When the direction of vector $\vec{Q}$ is reversed, the new angle between $\vec{P}$ and $-\vec{Q}$ becomes $(180^\circ – \theta)$. The new resultant is $\vec{S}$, so its magnitude is:

Since $\cos(180^\circ – \theta) = -\cos\theta$, we can rewrite this as:

Now, add Equation 1 and Equation 2:

Hence proved.

Let the two equal forces be $F_1 = F$ and $F_2 = F$. Let the angle between them be $\theta$.

The product of their magnitudes is $F \times F = F^2$.
We are given that the square of their resultant ($R^2$) is equal to three times their product:

According to the formula for the resultant of two vectors:

Substitute $F_1 = F, F_2 = F$, and $R^2 = 3F^2$ into the equation:

Since $\cos\theta = 0.5$, the angle is $\theta = 60^\circ$.

Let the two vectors have an equal magnitude $A$.
The angle between them is $\theta = \frac{2\pi}{3}$ radians, which is equal to $120^\circ$.

The magnitude of the resultant ($R$) is:

We know that $\cos(120^\circ) = -0.5$ (or $-\frac{1}{2}$). Substituting this value:

Therefore, the magnitude of the resultant is exactly equal to the magnitude of either individual vector. Hence proved.

Let the two forces be $F_1 = (P + Q)$ and $F_2 = (P – Q)$.
The resultant is $R = \sqrt{3P^2 + Q^2}$, which means $R^2 = 3P^2 + Q^2$.

Using the vector addition formula $R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$, we substitute our values:

Expand the squares using algebraic identities: $(P + Q)^2 + (P – Q)^2 = 2(P^2 + Q^2)$, and $(P + Q)(P – Q) = P^2 – Q^2$.

Subtract $2P^2 + 2Q^2$ from the left side:

Divide both sides by $(P^2 – Q^2)$ (assuming $P \neq Q$):

Therefore, the angle $\theta = 60^\circ$.