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Distance Formula

L1:x-12=y-23=...

Question

$L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$
$L_{2} : \frac{x - 2}{3} = \frac{y - 4}{2} = \frac{z - 5}{5}$ be two given lines, point P lies on $L_{1}$ and Q lies on $L_{2}$ then distance between P and Q can be

A

\frac{1}{3}

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B

\frac{1}{9}

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C

15

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D

30

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Solution

The correct options are
A $30$
C $15$
D $\frac{1}{3}$
$\frac{| ∣ ∣ ∣ ∣ \begin{matrix} 2 - 1 & 4 - 2 & 5 - 3 2 & 3 & 4 3 & 2 & 5 \end{matrix} ∣ ∣ ∣ ∣ |}{| ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 2 & 3 & 4 3 & 2 & 5 \end{matrix} ∣ ∣ ∣ ∣ |} = \frac{1}{\sqrt{78}}$

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Similar questions

L_{1}

L_{2}

be the lines whose equation are

\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}

respectively A and B are two points on

L_{1}

L_{2}

respectively such that AB is perpendicular both the lines

L_{1}

L_{2}

.Find points A, B and hence find shortest distance between lines

L_{1}

L_{2}

Consider the line

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

The shortest distance between

L_{1}

L_{2}

L_{1} : \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{1}

L_{2} : \frac{x}{1} = \frac{y}{- 1} = \frac{z - 5}{3}

The equation of the line perpendicular to

L_{1}

L_{2}

and passing through the point of intersection of

L_{1}

L_{2}

Q. Two lines whose equation are

L_{1} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{λ}

L_{2} : \frac{x - 2}{3} = \frac{y - 3}{2} = \frac{z - 2}{3}

lie in the same plane. If

L_{1}

intersects a plane

x + y + z = 15

at P, then distance of P from (3, 4, 3) is

Q. Find the shortest distance between the skew lines :

l_{1} : \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 2}{4}

l_{2} : \frac{x + 2}{4} = \frac{y - 0}{- 3} = \frac{z + 1}{1}

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