JEE Mathematics 2026 Question with Solution

Given: $\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$, $\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$, $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k}) = 4$.

Find: $|\vec{a}+\vec{c}|^2$.

Let

$$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$$

Using $\vec{a} \times \vec{c} = \vec{b}$,

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

So,

$$(2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k} = 8\hat{i} + 7\hat{j} - 3\hat{k}$$

Comparing components,

$$c_3 - c_2 = 4$$ $$c_3 + 2c_1 = 7$$ $$c_2 + 2c_1 = 3$$

From the dot product condition,

$$c_1 + c_2 + c_3 = 4$$

From $c_2 + 2c_1 = 3$, we get

$$c_2 = 3 - 2c_1$$

Using $c_3 - c_2 = 4$,

$$c_3 = 4 + c_2 = 4 + (3 - 2c_1) = 7 - 2c_1$$

Now substitute in $c_1 + c_2 + c_3 = 4$:

$$c_1 + (3 - 2c_1) + (7 - 2c_1) = 4$$ $$10 - 3c_1 = 4$$ $$3c_1 = 6 \implies c_1 = 2$$

Hence,

$$c_2 = 3 - 2(2) = -1$$ $$c_3 = 7 - 2(2) = 3$$

Therefore,

$$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$$

Now,

$$\vec{a}+\vec{c} = (-\hat{i} + 2\hat{j} + 2\hat{k}) + (2\hat{i} - \hat{j} + 3\hat{k}) = \hat{i} + \hat{j} + 5\hat{k}$$

Thus,

$$|\vec{a}+\vec{c}|^2 = 1^2 + 1^2 + 5^2 = 27$$

Therefore, the correct option is C.

Given: $\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$.

Find: the value of $|\vec{a}+\vec{c}|^2$.

Write $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then from the cross product condition we get three equations. However, one of them is dependent on the others, so the extra dot product condition is necessary to determine a unique vector.

The equations are:

$$c_3 - c_2 = 4$$ $$c_3 + 2c_1 = 7$$ $$c_2 + 2c_1 = 3$$

and

$$c_1 + c_2 + c_3 = 4$$

From the third equation,

$$c_2 = 3 - 2c_1$$

Then from the first equation,

$$c_3 = 4 + c_2 = 7 - 2c_1$$

Substituting these into the fourth equation,

$$c_1 + (3 - 2c_1) + (7 - 2c_1) = 4$$ $$10 - 3c_1 = 4$$ $$c_1 = 2$$

Therefore,

$$c_2 = -1, \qquad c_3 = 3$$

So,

$$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$$

Now add vectors:

$$\vec{a}+\vec{c} = (-1+2)\hat{i} + (2-1)\hat{j} + (2+3)\hat{k} = \hat{i} + \hat{j} + 5\hat{k}$$

Hence,

$$|\vec{a}+\vec{c}|^2 = 1 + 1 + 25 = 27$$

Therefore, the value is $27$, so the correct option is C.