Vector Algebra – Cross Product – JEE Main 21 Jan 2026 Shift 1

Solution:

Let $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.
Given $\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 8\hat{i} + 7\hat{j} – 3\hat{k}$.

Calculate $\vec{a} \times \vec{c}$:
$$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ c_1 & c_2 & c_3 \end{vmatrix}$$
$$= \hat{i}(2c_3 – 2c_2) – \hat{j}(-c_3 – 2c_1) + \hat{k}(-c_2 – 2c_1)$$
$$= (2c_3 – 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} – (c_2 + 2c_1)\hat{k}$$

Equating this to $\vec{b} = 8\hat{i} + 7\hat{j} – 3\hat{k}$:
1) $2c_3 – 2c_2 = 8 \Rightarrow c_3 – c_2 = 4$
2) $2c_1 + c_3 = 7$
3) $-(2c_1 + c_2) = -3 \Rightarrow 2c_1 + c_2 = 3$

We are also given $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$, which implies:
4) $c_1 + c_2 + c_3 = 4$

Substitute $c_3 = 7 – 2c_1$ and $c_2 = 3 – 2c_1$ into equation (4):
$$c_1 + (3 – 2c_1) + (7 – 2c_1) = 4$$
$$10 – 3c_1 = 4 \Rightarrow 3c_1 = 6 \Rightarrow c_1 = 2$$

Now find $c_2$ and $c_3$:
$$c_2 = 3 – 2(2) = -1$$
$$c_3 = 7 – 2(2) = 3$$
So, $\vec{c} = 2\hat{i} – \hat{j} + 3\hat{k}$.

We need to find $|\vec{a} + \vec{c}|^2$.
$$\vec{a} + \vec{c} = (-\hat{i} + 2\hat{j} + 2\hat{k}) + (2\hat{i} – \hat{j} + 3\hat{k})$$
$$\vec{a} + \vec{c} = \hat{i} + \hat{j} + 5\hat{k}$$

Magnitude squared:
$$|\vec{a} + \vec{c}|^2 = (1)^2 + (1)^2 + (5)^2$$
$$= 1 + 1 + 25 = 27$$

Ans. (2)