CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Shortest Distance between Two Skew Lines

Let L 1 an...

Question

Let $L_{1}$ and $L_{2}$ be the lines whose equation are $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ respectively A and B are two points on $L_{1}$ and $L_{2}$ respectively such that AB is perpendicular both the lines $L_{1}$ and $L_{2}$ .Find points A, B and hence find shortest distance between lines $L_{1}$ and $L_{2}$

Open in App

Solution

Given that,
$\begin{matrix} L_{1} : \frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} A (3 r_{1} + 3, - r_{1} + 8, r_{1} + 3) L_{2} : \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} B (- 3 r_{2} - 3, 2 r_{2} - 7, 4 r_{2} + 6) \end{matrix}$
Direction ratio of $P Q = - 3 r_{2} - 3 r_{1} - 6, 2 r_{2} + r_{1} - 15, 4 r_{2} - r_{1} + 3$
$\begin{matrix} - 3 r_{2} - 3 r_{1} - 6, 2 r_{2} + r_{1} - 15, 4 r_{2} - r_{1} + 3 - 3 (3 r_{1} + 3 r_{2} + 6) - 2 r_{2} - r_{1} + 15 + 4 r_{2} - r_{1} + 3 = 0 - 3 (3 r_{1} + 3 r_{2} + 6) + 2 (2 r_{2} + r_{1} - 15) + 4 (4 r_{2} - r_{1} + 3) = 0 - 11 r_{1} - 7 r_{2} = 0 7 r_{1} + 29 r_{2} = 0 r_{1} = 0 r_{2} = 0 A (3, 8, 3) B (- 3, - 7, 6) \to r_{1} = 3 ˆ i + 8 ˆ j + 3 ˆ k + λ (3 ˆ i - ˆ j + ˆ k) \to r_{2} = 3 ˆ i + 8 ˆ j + 3 ˆ k + λ (3 ˆ i - ˆ j + ˆ k) \end{matrix}$
Shortest distance $= \frac{∣ ∣ (6 ˆ i + 15 ˆ j - 3 ˆ k) . (3 ˆ i - ˆ j + ˆ k) \times (- 3 ˆ i + 2 ˆ j + 4 ˆ k) ∣ ∣}{∣ ∣ (3 ˆ i - ˆ j + ˆ k) \times (- 3 ˆ i + 2 ˆ j + 4 ˆ k) ∣ ∣}$
$\begin{matrix} = \frac{36 + 225 + 9}{\sqrt{36 + 225 + 9}} = \sqrt{270} \end{matrix}$
Then,
We get $L_{1}$ and $L_{2}$ is $\sqrt{270}$ .

Suggest Corrections

thumbs-up

0

Similar questions

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}

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}{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}

L_{2}

then distance between P and Q can be

Q. Consider the lines:

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 unit vector perpendicular to both

L_{1}

L_{2}

Q. Consider the lines

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 distance of the point

(1, 1, 1)

from the plane passing through the point

(- 1, - 2, - 1)

and whose normal is perpendicular to both lines

L_{1}

L_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Shortest Distance between Two Skew Lines

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Shortest Distance between Two Skew Lines

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon