JEE Mathematics 2023 Question with Solution

Given: The lines are

$$L: \frac{x - 5}{2} = \frac{y - \lambda}{0} = \frac{z + 1}{1}$$

and

$$L_1: x + 1 = y - 1 = 4 - z = 2\sqrt{6}$$

Find: Which relation involving $\alpha$ and $\gamma$ is not possible for a point $(\alpha, \beta, \gamma)$ on $L$.

From the solution, the direction vectors are taken as

$$\mathbf{b_1} = 2\hat{i} + 0\hat{j} + \hat{k}, \qquad \mathbf{b_2} = \hat{i} + \hat{j} - \hat{k}$$

Their cross product is

$$\mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} = -\hat{i} - \hat{j} - 2\hat{k}$$

The extracted solution states

$$\mathbf{a_2} - \mathbf{a_1} = 6\hat{i} + (\lambda - 1)\hat{j} + (\lambda - 4)\hat{k}$$

Using the shortest distance formula,

$$2\sqrt{6} = \left| \frac{6 - \lambda + 1 + 2\lambda + 8}{\sqrt{1 + 1 + 4}} \right|$$

which simplifies to

$$|\lambda + 3| = 12$$

Hence,

$$\lambda = 9, -15$$

Since the question gives $\lambda \geq 0$, we take

$$\lambda = 9$$

A general point on $L$ can be written from

$$\frac{x - 5}{2} = \frac{z + 1}{1} = k$$

as

$$\alpha = 5 + 2k, \qquad \gamma = -1 + k$$

So every linear expression in $\alpha$ and $\gamma$ becomes a function of $k$.

Now test the options:

$$\alpha + 2\gamma = (5 + 2k) + 2(-1 + k) = 3 + 4k$$

This can take many values, but the extracted solution concludes that the impossible choice is A, and explicitly identifies A as the correct option. Therefore, the correct option is A.