JEE Advanced 2019 — Paper 1 — Question 39

Equation of $L_1: \mathbf{r} = \mathbf{i} + \lambda(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ Equation of $L_2: \mathbf{r} = \mu(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})$

$L_1$ & $L_2$ are skew lines The direction ratios of line AB which is perpendicular to $L_1$ and $L_2$ will be

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 2 \\ 2 & -1 & 2 \end{vmatrix} = (4 - (-2))\mathbf{i} - (-2 - 4)\mathbf{j} + (1 - 4)\mathbf{k} = 6\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}$$ Hence direction ratios of AB will be $(2, 2, -1)$ direction ratios of AB proportional to $ (2, 2, -1)$

1 - \lambda - 2\mu = 2k \quad \ldots (i)

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2\lambda + \mu = 2k \quad \ldots (ii)

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2\lambda - 2\mu = -k \quad \ldots (iii)

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\text{solving } (i) \text{ & } (ii) \text{ & } (iii)

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\text{we get } \lambda = 1/9

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\mu = 2/9

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A\left(1 - \frac{1}{9}, 0 + 2 \times \frac{1}{9}, 0 + 2 \times \frac{1}{9}\right) = A\left(\frac{8}{9}, \frac{2}{9}, \frac{2}{9}\right)

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B\left(2 \times \frac{2}{9}, -1 \times \frac{2}{9}, 2 \times \frac{2}{9}\right) = B\left(\frac{4}{9}, -\frac{2}{9}, \frac{4}{9}\right)

Equation of line $ L_3 (A, B)$ passing through $ A$

\mathbf{r} = \left(\frac{8}{9}\mathbf{i} + \frac{2}{9}\mathbf{j} + \frac{2}{9}\mathbf{k}\right) + t(2\mathbf{i} + 2\mathbf{j} - \mathbf{k}), t \in \mathbb{R}

option (A) correct. Equation of line $ L_3$ passing through $B$

\mathbf{r} = \left(\frac{4}{9}\mathbf{i} - \frac{2}{9}\mathbf{j} + \frac{4}{9}\mathbf{k}\right) + t(2\mathbf{i} + 2\mathbf{j} - \mathbf{k}), t \in \mathbb{R}

$$Option (C) correct, option (B) also satisfy$$