JEE Mathematics 2025 Question with Solution

Given: $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, and the angle between them is $\frac{\pi}{3}$.

Find: The number of values of $\lambda$ in $[-1,3]$ for which $\lambda \mathbf{a}+2\mathbf{b}$ and $3\mathbf{a}-\lambda \mathbf{b}$ are perpendicular.

Since the vectors are perpendicular, their dot product is zero. Also,

$$\mathbf{a}\cdot\mathbf{b}=\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$

Now,

$$(\lambda \mathbf{a}+2\mathbf{b})\cdot(3\mathbf{a}-\lambda \mathbf{b})=0$$

Using distributivity,

$$3\lambda(\mathbf{a}\cdot\mathbf{a})-\lambda^2(\mathbf{a}\cdot\mathbf{b})+6(\mathbf{b}\cdot\mathbf{a})-2\lambda(\mathbf{b}\cdot\mathbf{b})=0$$

Since $\mathbf{a}$ and $\mathbf{b}$ are unit vectors,

$$\mathbf{a}\cdot\mathbf{a}=1,\quad \mathbf{b}\cdot\mathbf{b}=1,\quad \mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}=\frac{1}{2}$$

So,

$$3\lambda-\frac{\lambda^2}{2}+3-2\lambda=0$$ $$-\frac{\lambda^2}{2}+\lambda+3=0$$ $$\lambda^2-2\lambda-6=0$$

Hence,

$$\lambda=1\pm \sqrt{7}$$

Now check the interval $[-1,3]$:

• $1+\sqrt{7}>3$
• $1-\sqrt{7}<-1$ So no value of $\lambda$ lies in the given interval.

Therefore, the number of values of $\lambda$ is $0$, so the correct option is B.

The solution states $\lambda=0$ and then concludes there is only $1$ value, which is inconsistent with the full dot-product expansion above. The option marked in the solution, B, matches the correct count.

Given: $\angle(\mathbf{a},\mathbf{b})=\frac{\pi}{3}$ and both are unit vectors.

Find: How many solutions of $\lambda$ belong to $[-1,3]$.

First compute

$$\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\cos\frac{\pi}{3}=1\cdot 1\cdot \frac{1}{2}=\frac{1}{2}$$

Perpendicular vectors satisfy zero dot product:

$$(\lambda \mathbf{a}+2\mathbf{b})\cdot(3\mathbf{a}-\lambda \mathbf{b})=0$$

Expand term by term:

$$(\lambda \mathbf{a})\cdot(3\mathbf{a})+(\lambda \mathbf{a})\cdot(-\lambda \mathbf{b})+(2\mathbf{b})\cdot(3\mathbf{a})+(2\mathbf{b})\cdot(-\lambda \mathbf{b})=0$$ $$3\lambda(\mathbf{a}\cdot\mathbf{a})-\lambda^2(\mathbf{a}\cdot\mathbf{b})+6(\mathbf{b}\cdot\mathbf{a})-2\lambda(\mathbf{b}\cdot\mathbf{b})=0$$

Substitute the known dot products:

$$3\lambda-\lambda^2\left(\frac{1}{2}\right)+6\left(\frac{1}{2}\right)-2\lambda=0$$ $$3\lambda-\frac{\lambda^2}{2}+3-2\lambda=0$$ $$-\frac{\lambda^2}{2}+\lambda+3=0$$

Multiply by $-2$:

$$\lambda^2-2\lambda-6=0$$

Solve the quadratic:

$$\lambda=\frac{2\pm \sqrt{4+24}}{2}=\frac{2\pm \sqrt{28}}{2}=1\pm \sqrt{7}$$

Neither root lies in $[-1,3]$. Hence the required number of values is $0$.

Therefore, the correct option is B.