Let the line of the shortest distance between the lines and intersect and at P and Q respectively. If is the mid point of the line segment PQ, then is equal to [2024]

Question

Let the line of the shortest distance between the lines

$L_{1} : \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ and $L_{2} : \vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$

intersect $L_{1}$ and $L_{2}$ at P and Q respectively. If $(α, β, γ)$ is the mid point of the line segment PQ, then $2 (α + β + γ)$ is equal to __________. [2024]

Smart Achievers · Accepted Answer

(21)

We have, $L_{1} : \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$

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

Here, ${\vec{b}}_{1} = \hat{i} - \hat{j} + \hat{k}, {\vec{b}}_{2} = \hat{i} + \hat{j} - \hat{k}$

${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix} |$ $= 0 \hat{i} + 2 \hat{j} + 2 \hat{k}$

Direction ratios of line perpendicular to $L_{1}$ and $L_{2}$

$= ((4 + μ) - (1 + λ), (5 + μ) - (2 - λ), (6 - μ) - (3 + λ))$

$= 3 + μ - λ, 3 + μ + λ, 3 - μ - λ$

$∴ \frac{3 + μ - λ}{0} = \frac{3 + μ + λ}{2} = \frac{3 - μ - λ}{2}$

On solving, we get

$λ = \frac{3}{2}, μ = \frac{- 3}{2}$

Point $P \equiv (\frac{5}{2}, \frac{1}{2}, \frac{9}{2})$ and $Q \equiv (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})$

$∴$ Mid point of AB = $(\frac{5}{2}, 2, 6) = (α, β, γ)$

$∴ 2 (α + β + γ)$ = 5 + 4 + 12 = 21.

Solution

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Solution

Q. Let the line of the shortest distance between the lines L1:r→=(i^+2j^+3k^)+λ(i^–j^+k^) and L2:r→=(4i^+5j^+6k^)+μ(i^+j^–k^) intersect L1 and L2 at P and Q respectively. If (α,β,γ) is the mid point of the line segment PQ, then 2(α+β+γ) is equal to __________. [2024]

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