Vector Algebra – Vector Product – JEE Main 22 Jan 2026 Shift 2

Solution:

First, we find $\vec{c} = \vec{a} \times \vec{b}$.
$$ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & \lambda & 2 \end{vmatrix} $$
$$ \vec{c} = \hat{i}(-2 – \lambda) – \hat{j}(4 – 0) + \hat{k}(2\lambda – 0) $$
$$ \vec{c} = (-2 – \lambda)\hat{i} – 4\hat{j} + 2\lambda\hat{k} $$

Given $|\vec{c}| = \sqrt{53}$:
$$ (-2 – \lambda)^2 + (-4)^2 + (2\lambda)^2 = 53 $$
$$ (4 + \lambda^2 + 4\lambda) + 16 + 4\lambda^2 = 53 $$
$$ 5\lambda^2 + 4\lambda + 20 = 53 $$
$$ 5\lambda^2 + 4\lambda – 33 = 0 $$
$$ (\lambda + 3)(5\lambda – 11) = 0 $$
Since $\lambda \in Z$, we get $\lambda = -3$.

Substitute $\lambda = -3$ back into $\vec{c}$:
$$ \vec{c} = ( -2 – (-3) )\hat{i} – 4\hat{j} + 2(-3)\hat{k} $$
$$ \vec{c} = 1\hat{i} – 4\hat{j} – 6\hat{k} $$

Now, about vector $\vec{d}$. It is in the yz-plane and has magnitude 2.
Let $\vec{d} = 0\hat{i} + y\hat{j} + z\hat{k}$.
Since $|\vec{d}| = 2$, we have $y^2 + z^2 = 4$.

We need to maximize the dot product squared: $(\vec{c} \cdot \vec{d})^2$.
$$ \vec{c} \cdot \vec{d} = (1)(0) + (-4)(y) + (-6)(z) = -4y – 6z $$
Let’s use a simple formula:
For any expression $Ay + Bz$, the maximum value squared is $(A^2 + B^2)(y^2 + z^2)$.

Here $A = -4$ and $B = -6$.
$$ \text{Max } (-4y – 6z)^2 = [(-4)^2 + (-6)^2] [y^2 + z^2] $$
$$ = [16 + 36] [4] $$
$$ = [52] [4] $$
$$ = 208 $$

Ans. (3)