First slide

Straight lines in 3D

Question

The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1} a n d x + y + z + 1 = 0, 2 x - y + z + 3 = 0 i s :$

Moderate

A

1
B

$\frac{1}{\sqrt{3}}$
C

$\frac{1}{\sqrt{2}}$
D

$\frac{1}{2}$

Solution

$\begin{array}{l} L_{1} : \frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1} \\ L_{2} : x + y + z + 1 = 0, 2 x - y + z + 3 = 0 \\ Equation of plane (π = 0) through L_{2} and parallel to L_{1} is \\ x + y + z + 1 + λ (2 x - y + z + 3) = 0 \\ 0 (1 + 2 λ) - 1 (1 - λ) + 1 (1 + λ) = 0 \\ - 1 + λ + 1 + λ = 0 \\ λ = 0 \\ x + y + z + 1 = 0 \end{array}$
$Shortest distance = distance of (1, - 1, 0) from x + y + z + 1 = 0 is \frac{|p^{'} - γ + {}^{2}+ 1|}{\sqrt{1 + 1 + 1}} = \frac{1}{\sqrt{3}}$

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