Let $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ denotes the lines $$\begin{aligned} & \vec{r}=\hat{i}+\lambda(-\hat{i}+2 \hat{j}+2 \hat{k}), \lambda \in \mathbb{R} \text { and } \\ & \vec{r}=\mu(2 \hat{i}-\hat{j}+2 \hat{k}), \mu \in \mathbb{R} \end{aligned}$$ respectively. If $\mathrm{L}_{3}$ is a line which is perpendicular to both $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ and cuts both of them, then which of the following options describe(s) $\mathrm{L}_{3}$ ?

(A) $\overrightarrow{\mathrm{r}}=\frac{1}{3}(2 \hat{\mathrm{i}}+\hat{\mathrm{k}})+\mathrm{t}(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}), \mathrm{t} \in \mathbb{R}$; (B) $\overrightarrow{\mathrm{r}}=\frac{2}{9}(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mathrm{t}(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}), \mathrm{t} \in \mathbb{R}$; (D) $\overrightarrow{\mathrm{r}}=\frac{2}{9}(4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{t}(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}), \mathrm{t} \in \mathbb{R}$

(C) $\overrightarrow{\mathrm{r}}=\mathrm{t}(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}), \mathrm{t} \in \mathbb{R}$

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Three Dimensional Geometry

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Let

L_{1}

and

L_{2}

denotes the lines

r = \hat{i} + λ (- \hat{i} + 2 \hat{j} + 2 \hat{k}), λ \in R and r = μ (2 \hat{i} - \hat{j} + 2 \hat{k}), μ \in R

respectively. If

L_{3}

is a line which is perpendicular to both

L_{1}

and

L_{2}

and cuts both of them, then which of the following options describe(s)

L_{3}

(A) $r = \frac{1}{3} (2 \hat{i} + \hat{k}) + t (2 \hat{i} + 2 \hat{j} - \hat{k}), t \in R$ ✓ Correct
(B) $r = \frac{2}{9} (2 \hat{i} - \hat{j} + 2 \hat{k}) + t (2 \hat{i} + 2 \hat{j} - \hat{k}), t \in R$ ✓ Correct
(C) $r = t (2 \hat{i} + 2 \hat{j} - \hat{k}), t \in R$
(D) $r = \frac{2}{9} (4 \hat{i} + \hat{j} + \hat{k}) + t (2 \hat{i} + 2 \hat{j} - \hat{k}), t \in R$ ✓ Correct

Answer: (A), (B), (D)

Solution

Points on

L_{1}

and

L_{2}

are

A (1 - λ, 2 λ, 2 λ)

and

B (2 μ, - μ, 2 μ)

respectively.

The direction of the shortest distance is

d = (- \hat{i} + 2 \hat{j} + 2 \hat{k}) \times (2 \hat{i} - \hat{j} + 2 \hat{k}) = 2 \hat{i} + 2 \hat{j} - \hat{k}

.

For

AB

to be parallel to

d

\frac{2 μ + λ - 1}{2} = \frac{- μ - 2 λ}{2} = \frac{2 μ - 2 λ}{- 1}

Solving:

λ = \frac{1}{9}

and

μ = \frac{2}{9}

.

This gives

A \equiv (\frac{8}{9}, \frac{2}{9}, \frac{2}{9})

and

B \equiv (\frac{4}{9}, - \frac{2}{9}, \frac{4}{9})

.

The line

L_{3}

passes through both A and B with direction

2 \hat{i} + 2 \hat{j} - \hat{k}

. Options (A), (B), and (D) all represent this line (passing through A, B, or their midpoint respectively).

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Three Dimensional Geometry

Solution

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