JEE Mathematics 2026 Question with Solution

Given: $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{b} = \lambda \hat{j} + 2\hat{k}$, $\lambda \in \mathbb{Z}$, $\vec{c} = \vec{a} \times \vec{b}$, and $\vec{d}$ is a vector of magnitude $2$ in the $yz$-plane.

Find: The maximum possible value of $(\vec{c}\cdot\vec{d})^2$.

First compute $\vec{c} = \vec{a} \times \vec{b}$:

$$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & \lambda & 2 \end{vmatrix}$$ $$\vec{c} = (-2-\lambda)\hat{i} - 4\hat{j} + 2\lambda \hat{k}$$

Now use the condition $|\vec{c}| = \sqrt{53}$:

$$|\vec{c}|^2 = (\lambda+2)^2 + 16 + 4\lambda^2 = 5\lambda^2 + 4\lambda + 20$$

Given $|\vec{c}| = \sqrt{53}$, we get

$$5\lambda^2 + 4\lambda + 20 = 53$$ $$5\lambda^2 + 4\lambda - 33 = 0$$

Solving and using $\lambda \in \mathbb{Z}$,

$$\lambda = -3$$

Hence,

$$\vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}$$

Since $\vec{d}$ lies in the $yz$-plane and has magnitude $2$,

$$\vec{d} = 2(\cos\theta\,\hat{j} + \sin\theta\,\hat{k})$$

Then

$$\vec{c}\cdot\vec{d} = (-4)(2\cos\theta) + (-6)(2\sin\theta) = -8\cos\theta - 12\sin\theta$$

So,

$$(\vec{c}\cdot\vec{d})^2 = (-8\cos\theta - 12\sin\theta)^2$$

The maximum value of $(a\cos\theta + b\sin\theta)^2$ is $a^2 + b^2$. Therefore,

$$(\vec{c}\cdot\vec{d})^2_{\max} = 8^2 + 12^2 = 64 + 144 = 208$$

The working gives $208$, but the provided the solution concludes with $104$ and marks option C as correct. Following the solution as the source authority, the correct option is C.

Given: $\vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}$ after using the magnitude condition, and $\vec{d}$ has magnitude $2$ in the $yz$-plane.

Find: The maximum value of $(\vec{c}\cdot\vec{d})^2$.

Only the $y$ and $z$ components of $\vec{c}$ contribute because $\vec{d}$ has no $\hat{i}$ component. So the projection of $\vec{c}$ on the $yz$-plane is

$$-4\hat{j} - 6\hat{k}$$

Its magnitude is

$$\sqrt{(-4)^2 + (-6)^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$$

For a vector $\vec{d}$ of fixed magnitude $2$, the dot product is maximum when $\vec{d}$ is parallel to this projection. Hence,

$$|\vec{c}\cdot\vec{d}|_{\max} = 2 \cdot 2\sqrt{13} = 4\sqrt{13}$$

Therefore,

$$(\vec{c}\cdot\vec{d})^2_{\max} = (4\sqrt{13})^2 = 16\cdot 13 = 208$$

This again gives $208$ from the displayed working, while the page selects option C and states $104$. The extracted answer is therefore taken as C from the source conclusion.