If the foot of the perpendicular from point 4- 3- 8 on the line L1 - x-al - y-23 - z-b4-l-0 is 3- 5- 7- then the shortest distance between the line L1 and line L2- x-23 - y-44 - z-55 is equal to -

(3, 5, 7) lies on given line L_1. Then 3 al = 33 = 7 b4 7 b4 = 1 nbsp; ⇒ b = 3 3 al = 1 ⇒ 3 a = l A(4, 3, 8) and B(3, 5, 7) DR's of AB is (1, 2, 1) AB ⊥ line ...

201

You visited us 201 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

If the foot o...

Question

If the foot of the perpendicular from point $(4, 3, 8)$ on the line $L_{1} : \frac{x - a}{l} = \frac{y - 2}{3} = \frac{z - b}{4},$
$l \neq 0$ is $(3, 5, 7),$ then the shortest distance between the line $L_{1}$ and line $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is equal to :

\sqrt{\frac{2}{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{\sqrt{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{\sqrt{6}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{1}{\sqrt{6}}$
$(3, 5, 7)$ lies on given line $L_{1} .$
Then $\frac{3 - a}{l} = \frac{3}{3} = \frac{7 - b}{4}$
$\frac{7 - b}{4} = 1 \Rightarrow b = 3$
$\frac{3 - a}{l} = 1 \Rightarrow 3 - a = l$
$A (4, 3, 8)$ and $B (3, 5, 7)$
DR's of $A B$ is $(1, - 2, 1)$
AB $⊥$ line $L_{1}$
$\Rightarrow (1) (l) + (- 2) (- 3) + 4 (1) = 0$
$\Rightarrow l = 2$
$∴ a = 1$

$a = 1, b = 3, l = 2$
$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$
$\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$
$Shortest distance = \frac{∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 2 2 & 3 & 4 3 & 4 & 5 \end{matrix} ∣ ∣ ∣ ∣}{∥ ∥ ∥ ∥ ∥ \begin{matrix} ^i & ^j & ^k 2 & 3 & 4 3 & 4 & 5 \end{matrix} ∥ ∥ ∥ ∥ ∥} = \frac{1}{\sqrt{6}}$

Suggest Corrections

Similar questions

Q. 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}

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

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}

Q. The equation of the line of shortest distance between the lines

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

and

\frac{x - 5}{5} = \frac{y - 3}{3} = \frac{z}{0}

, is

Q. The shortest distance between the lines

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

and

\frac{x - 8}{4} = \frac{y - 7}{- 5} = \frac{z - 5}{- 5}

Join BYJU'S Learning Program

If the foot of the perpendicular from point (4,3,8) on the line L1:x−al=y−23=z−b4, l≠0 is (3,5,7), then the shortest distance between the line L1 and line L2:x−23=y−44=z−55 is equal to :

If the foot of the perpendicular from point $(4, 3, 8)$ on the line $L_{1} : \frac{x - a}{l} = \frac{y - 2}{3} = \frac{z - b}{4},$
$l \neq 0$ is $(3, 5, 7),$ then the shortest distance between the line $L_{1}$ and line $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is equal to :