Let the line of the shortest distance between the lines \(L_{1}: \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(L_{2}: \vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})\) intersect \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\) at P and Q respectively. If \((\alpha, \beta, \gamma)\) is the midpoint of the line segment PQ, then \(2(\alpha+\beta+\gamma)\) is equal to ______.

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Three Dimensional Geometry

Let the line of the shortest distance between the lines

L_{1} : r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})

and

L_{2} : r = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})

intersect

L_{1}

and

L_{2}

at P and Q respectively. If

(α, β, γ)

is the midpoint of the line segment PQ, then

2 (α + β + γ)

is equal to ______.

Answer: 21

L_{1} : r = (1, 2, 3) + λ (1, - 1, 1)

L_{2} : r = (4, 5, 6) + μ (1, 1, - 1)

.
The common perpendicular direction is

n = (1, - 1, 1) \times (1, 1, - 1) = ((- 1) (- 1) - (1) (1), (1) (1) - (1) (- 1), (1) (1) - (- 1) (1)) = (0, 2, 2)

, i.e. along

(0, 1, 1)

.
Let

P = (1 + λ, 2 - λ, 3 + λ)

L_{1}

and

Q = (4 + μ, 5 + μ, 6 - μ)

L_{2}

;

P Q

must be parallel to

(0, 1, 1)

P Q = (3 + μ - λ, 3 + μ + λ, 3 - μ - λ)

x

-component zero:

3 + μ - λ = 0 \Rightarrow μ = λ - 3

y

-component

=

z

-component:

3 + μ + λ = 3 - μ - λ \Rightarrow μ = - λ

.
Solving together:

- λ = λ - 3 \Rightarrow λ = \frac{3}{2}, μ = - \frac{3}{2}

P = (\frac{5}{2}, \frac{1}{2}, \frac{9}{2}), Q = (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})

Midpoint:

(α, β, γ) = (\frac{5}{2}, 2, 6)

2 (α + β + γ) = 2 (\frac{5}{2} + 2 + 6) = 2 \times \frac{21}{2} = 21

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer