JEE Mathematics 2026 Question with Solution

Given: $\vec a = 2\hat i + \hat j - 2\hat k$, $\vec b = \hat i + \hat j$, $\vec c = \vec a \times \vec b$, $|\vec d - \vec a| = \sqrt{11}$, $|\vec c \times \vec d| = 3$, and the angle between $\vec c$ and $\vec d$ is $\frac{\pi}{4}$.

Find: $\vec a \cdot \vec d$.

First compute $\vec c$:

$$\vec c = \vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix}$$ $$\vec c = 2\hat i - 2\hat j + \hat k$$

Now use the magnitude of cross product:

$$|\vec c \times \vec d| = |\vec c||\vec d|\sin\theta$$ $$3 = |\vec c||\vec d|\sin\frac{\pi}{4}$$

Also,

$$|\vec c| = \sqrt{4+4+1} = 3$$

So,

$$3 = 3|\vec d|\cdot \frac{1}{\sqrt{2}}$$

which gives

$$|\vec d| = \sqrt{2}$$

Now use

$$|\vec d-\vec a|^2 = |\vec d|^2 + |\vec a|^2 - 2\vec a\cdot\vec d$$

Since $|\vec d-\vec a| = \sqrt{11}$, $|\vec d|^2 = 2$, and

$$|\vec a|^2 = 2^2 + 1^2 + (-2)^2 = 9$$

we get

$$11 = 2 + 9 - 2\vec a\cdot\vec d$$ $$11 = 11 - 2\vec a\cdot\vec d$$ $$\vec a\cdot\vec d = 0$$

Therefore, $\vec a \cdot \vec d = 0$, so the correct option is D.

Given: $\vec a = 2\hat i + \hat j - 2\hat k$, $\vec b = \hat i + \hat j$.

Find: the value of $\vec a \cdot \vec d$.

From the determinant expansion provided in the solution,

$$\vec c = \vec a \times \vec b = 2\hat i - 2\hat j + \hat k$$

Hence,

$$|\vec c| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$$

The magnitude condition gives

$$|\vec c \times \vec d| = |\vec c||\vec d|\sin\theta$$

with $\theta = \frac{\pi}{4}$. Therefore,

$$3 = 3|\vec d|\sin\frac{\pi}{4}$$ $$3 = 3|\vec d|\cdot \frac{1}{\sqrt{2}}$$ $$|\vec d| = \sqrt{2}$$

Now square the given distance condition:

$$|\vec d - \vec a| = \sqrt{11} \Rightarrow |\vec d - \vec a|^2 = 11$$

Using the identity,

$$|\vec d - \vec a|^2 = |\vec d|^2 + |\vec a|^2 - 2\vec a \cdot \vec d$$

we need $|\vec a|^2$:

$$|\vec a|^2 = 2^2 + 1^2 + (-2)^2 = 9$$

Substituting,

$$11 = 2 + 9 - 2\vec a \cdot \vec d$$ $$11 = 11 - 2\vec a \cdot \vec d$$ $$2\vec a \cdot \vec d = 0$$ $$\vec a \cdot \vec d = 0$$

Therefore, the correct option is D.