JEE Mathematics 2025 Question with Solution

Given: $\vec{b} = \lambda \hat{i} + 4 \hat{k}$, $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$, and $\vec{c}$ is the projection of $\vec{b}$ on $\vec{a}$. Also, $|\vec{a} + \vec{c}| = 7$.

Find: The area of the parallelogram formed by $\vec{b}$ and $\vec{c}$.

The projection of $\vec{b}$ on $\vec{a}$ is

$$\vec{c} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \, \vec{a}$$

Now,

$$\vec{a} \cdot \vec{b} = (\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (\lambda \hat{i} + 4\hat{k}) = \lambda + 8$$

and

$$|\vec{a}|^2 = 1^2 + 2^2 + 2^2 = 9$$

Hence,

$$\vec{c} = \frac{\lambda + 8}{9}(\hat{i} + 2\hat{j} + 2\hat{k})$$

So,

$$\vec{a} + \vec{c} = \left(1+\frac{\lambda+8}{9}\right)\hat{i} + \left(2+\frac{2(\lambda+8)}{9}\right)\hat{j} + \left(2+\frac{2(\lambda+8)}{9}\right)\hat{k}$$

which simplifies to

$$\vec{a} + \vec{c} = \frac{\lambda+17}{9}\hat{i} + \frac{2(\lambda+17)}{9}\hat{j} + \frac{2(\lambda+17)}{9}\hat{k}$$

Therefore,

$$|\vec{a}+\vec{c}| = \frac{|\lambda+17|}{9}\sqrt{1^2+2^2+2^2} = \frac{|\lambda+17|}{3}$$

Given that $|\vec{a}+\vec{c}| = 7$,

$$\frac{|\lambda+17|}{3} = 7$$

So,

$$|\lambda+17| = 21$$

Since $\lambda > 0$, we get

$$\lambda = 4$$

Now,

$$\vec{b} = 4\hat{i} + 4\hat{k}$$

and

$$\vec{c} = \frac{4+8}{9}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j} + \frac{8}{3}\hat{k}$$

The area of the parallelogram is

$$|\vec{b} \times \vec{c}|$$

Now,

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & 4 \\ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \end{vmatrix}$$

Expanding,

$$\vec{b} \times \vec{c} = \frac{1}{3}(-32\hat{i} - 16\hat{j} + 32\hat{k})$$

Thus,

$$|\vec{b} \times \vec{c}| = \frac{1}{3}\sqrt{(-32)^2 + (-16)^2 + 32^2} = \frac{1}{3}\sqrt{2304} = 16$$

Therefore, the area of the parallelogram is $16$.

Given: the solution concludes that the area is $16$.

Find: The numerical value of the area.

The first approach states $\lambda = 2$, but the detailed working in the second approach shows

$$|\vec{a}+\vec{c}| = \frac{|\lambda+17|}{3} = 7$$

which gives

$$|\lambda+17| = 21$$

and since $\lambda > 0$,

$$\lambda = 4$$

Using $\lambda = 4$,

$$\vec{b} = 4\hat{i} + 4\hat{k}, \qquad \vec{c} = \frac{4}{3}\hat{i} + \frac{8}{3}\hat{j} + \frac{8}{3}\hat{k}$$

Then,

$$\vec{b} \times \vec{c} = \frac{1}{3}(-32\hat{i} - 16\hat{j} + 32\hat{k})$$

and hence

$$|\vec{b} \times \vec{c}| = 16$$

So the correct extracted answer is $16$. The intermediate statement $\lambda = 2$ in the first approach is inconsistent with the detailed algebra and with the final answer.