JEE Mathematics 2023 Question with Solution

Given:

• $\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$
• $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$
• $$$$

(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0

- $$\vec{a} \cdot \vec{c} = -17$$ - $$\vec{b} \cdot \vec{c} = -20$$

Find: $\left| \vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k}) \right|^2$

From

$$(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = 0$$

we get

$$(\vec{a} + \vec{b}) \times \vec{c} = 0.$$

Hence $\vec{c}$ is parallel to $\vec{a} + \vec{b}$, so let

$$\vec{c} = \alpha(\vec{a} + \vec{b}) = \alpha(\lambda + 3)\hat{i} + \alpha \hat{k}.$$

Using $\vec{b} \cdot \vec{c} = -20$,

$$3\alpha(\lambda + 3) + 2\alpha = -20.$$

Using $\vec{a} \cdot \vec{c} = -17$,

$$\alpha(\lambda(\lambda + 3)) - \alpha = -17.$$

So,

$$\alpha(3\lambda + 11) = -20$$

and

$$\alpha(\lambda^2 + 3\lambda - 1) = -17.$$

Eliminating $\alpha$,

$$17(3\lambda + 11) = 20(\lambda^2 + 3\lambda - 1).$$

Thus,

$$20\lambda^2 + 9\lambda - 207 = 0.$$

Since $\lambda \in \mathbb{Z}$,

$$\lambda = 3.$$

Substituting, the solution gives

$$\alpha = -1, \quad \vec{c} = -6\hat{i} + \hat{k}.$$

Now compute

$$\vec{v} = \vec{c} \times (3\hat{i} + \hat{j} + \hat{k}).$$

Using the determinant,

$$\vec{v} = \hat{i} + 3\hat{j} - 6\hat{k}.$$

Therefore,

$$|\vec{v}|^2 = 1^2 + 3^2 + (-6)^2 = 46.$$

So the computed value is $46$. However, the solution explicitly states "The Correct Option is A", while option A = $62$ and option B = $46$. Following the instruction that the solution is the primary source for the answer label, the answer is recorded as A, though the numerical working matches option B.