First slide

Straight lines in 3D

Question

The equation of line of shortest distance between the lines $\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1};$ $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$ is

Moderate

A

$\frac{x - 3}{2} = \frac{y - 5}{3} = \frac{z - 7}{4}$
B

$\frac{x + 3}{2} = \frac{y + 5}{3} = \frac{z - 7}{4}$
C

$\frac{x + 3}{2} = \frac{y - 5}{3} = \frac{z - 7}{4}$
D

$\frac{x - 3}{2} = \frac{y - 5}{3} = \frac{z + 7}{4}$

Solution

Let $P (α + 3, - 2 α + 5, α + 7)$ and $Q (7 β - 1, - 6 β - 1, β - 1)$ be the points on the given lines so that PQ is the line of shortest distance between the given lines
D.r’s of $P Q = (α - 7 β + 4, - 2 α + 6 β + 6, α - β + 8)$
Since PQ is perpendicular to the given lines $1 (α - 7 β + 4) - 2 (- 2 α + 6 β + 6) + 1 (α - β + 8) = 0$
$\Rightarrow 6 α - 20 β = 0 \Rightarrow 3 α - 10 β = 0$ --- (1)
and,
$\Rightarrow$ $7 (α - 7 β + 4) - 6 (- 2 α + 6 β + 6) + 1 (α - β + 8) = 0$
$\Rightarrow$ $7 α - 49 β + 28 + 12 α - 36 β - 36 + α - β + 8 = 0$
$\Rightarrow$ $20 α - 86 β = 0 \Rightarrow 10 α = 43 β$ --- (2)
From (1) and (2) $α = 0, β = 0$
$P = (3, 5, 7), Q = (- 1, - 1, - 1)$
D.r’s of PQ = (–4, –6, –8) = (2, 3, 4)
Equation of PQ is $\frac{x - 3}{2} = \frac{y - 5}{3} = \frac{z - 7}{4}$

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