JEE Mathematics 2025 Question with Solution

Given: $\mathbf{a} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\mathbf{b} = 3\hat{i} - 5\hat{j} + \hat{k}$, $\mathbf{a} \times \mathbf{c} = \mathbf{a} \times \mathbf{b}$, and $(\mathbf{a} + \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c}) = 168$.

Find: The maximum value of $|\mathbf{c}|^2$.

From

$$\mathbf{a} \times \mathbf{c} = \mathbf{a} \times \mathbf{b}$$

we get

$$\mathbf{a} \times (\mathbf{c} - \mathbf{b}) = 0$$

so $\mathbf{c} - \mathbf{b}$ is parallel to $\mathbf{a}$. Hence,

$$\mathbf{c} = \mathbf{b} + \lambda \mathbf{a}$$

for some scalar $\lambda$.

Now,

$$\mathbf{a} + \mathbf{c} = \mathbf{a} + \mathbf{b} + \lambda \mathbf{a} = (1+\lambda)\mathbf{a} + \mathbf{b}$$

and

$$\mathbf{b} + \mathbf{c} = 2\mathbf{b} + \lambda \mathbf{a}$$

Using the given dot product condition,

$$((1+\lambda)\mathbf{a} + \mathbf{b}) \cdot (2\mathbf{b} + \lambda \mathbf{a}) = 168$$

First compute:

$$\mathbf{a} \cdot \mathbf{a} = 2^2 + (-1)^2 + 3^2 = 14$$ $$\mathbf{b} \cdot \mathbf{b} = 3^2 + (-5)^2 + 1^2 = 35$$ $$\mathbf{a} \cdot \mathbf{b} = 2\cdot 3 + (-1)(-5) + 3\cdot 1 = 14$$

Substituting,

$$((1+\lambda)\mathbf{a} + \mathbf{b}) \cdot (2\mathbf{b} + \lambda \mathbf{a}) = 2(1+\lambda)(\mathbf{a}\cdot\mathbf{b}) + \lambda(1+\lambda)(\mathbf{a}\cdot\mathbf{a}) + 2(\mathbf{b}\cdot\mathbf{b}) + \lambda(\mathbf{b}\cdot\mathbf{a})$$ $$= 28(1+\lambda) + 14\lambda(1+\lambda) + 70 + 14\lambda$$ $$= 98 + 56\lambda + 14\lambda^2$$

Therefore,

$$98 + 56\lambda + 14\lambda^2 = 168$$ $$14\lambda^2 + 56\lambda - 70 = 0$$ $$\lambda^2 + 4\lambda - 5 = 0$$ $$(\lambda-1)(\lambda+5) = 0$$

So,

$$\lambda = 1 \quad \text{or} \quad \lambda = -5$$

Now,

$$\mathbf{c} = \mathbf{b} + \lambda \mathbf{a}$$

If $\lambda = 1$,

$$\mathbf{c} = \mathbf{a} + \mathbf{b} = 5\hat{i} - 6\hat{j} + 4\hat{k}$$

so

$$|\mathbf{c}|^2 = 5^2 + (-6)^2 + 4^2 = 77$$

If $\lambda = -5$,

$$\mathbf{c} = \mathbf{b} - 5\mathbf{a} = -7\hat{i} + 0\hat{j} - 14\hat{k}$$

so

$$|\mathbf{c}|^2 = (-7)^2 + 0^2 + (-14)^2 = 49 + 196 = 245$$

Hence the maximum value obtained from the given condition is $245$.

The solution concludes with option C, that is $308$, but this does not match the working implied by the stated question condition $\mathbf{a} \times \mathbf{c} = \mathbf{a} \times \mathbf{b}$. Since the solution declares C as correct, the marked answer is C.

Given: $\mathbf{a} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\mathbf{b} = 3\hat{i} - 5\hat{j} + \hat{k}$.

Find: The maximum value of $|\mathbf{c}|^2$.

The source solution rewrites the cross-product condition as

$$\mathbf{a} \times \mathbf{c} = \mathbf{c} \times \mathbf{b}$$

which gives

$$\mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c} = 0$$ $$(\mathbf{a} + \mathbf{b}) \times \mathbf{c} = 0$$

Hence $\mathbf{c}$ is parallel to $\mathbf{a} + \mathbf{b}$, so let

$$\mathbf{c} = \lambda(\mathbf{a} + \mathbf{b}) = \lambda(5\hat{i} - 6\hat{j} + 4\hat{k})$$

Then

$$|\mathbf{c}|^2 = \lambda^2(5^2 + (-6)^2 + 4^2) = 77\lambda^2$$

Also,

$$(\mathbf{a} + \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c}) = 168$$

Expanding,

$$\mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{b} + |\mathbf{c}|^2 = 168$$

Now,

$$\mathbf{a}\cdot\mathbf{b} = 14$$ $$\mathbf{a}\cdot(5\hat{i} - 6\hat{j} + 4\hat{k}) = 28$$ $$\mathbf{b}\cdot(5\hat{i} - 6\hat{j} + 4\hat{k}) = 49$$

So,

$$14 + 28\lambda + 49\lambda + 77\lambda^2 = 168$$ $$77\lambda^2 + 77\lambda - 154 = 0$$ $$\lambda^2 + \lambda - 2 = 0$$ $$(\lambda+2)(\lambda-1) = 0$$

Thus,

$$\lambda = -2 \quad \text{or} \quad \lambda = 1$$

Therefore,

$$|\mathbf{c}|^2 = 77\lambda^2$$

so the two values are $308$ and $77$. The maximum is $308$.

This matches the recorded correct option C. However, this approach is based on a cross-product condition different from the one printed in the question.