Consider the line L passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line L along the line $\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$ is equal to :

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Three Dimensional Geometry

Consider the line L passing through the points

(1, 2, 3)

and

(2, 3, 5)

. The distance of the point

(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})

from the line L along the line

\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}

is equal to :

Answer: (A)

Line L through

(1, 2, 3)

and

(2, 3, 5)

has direction ratios

(1, 1, 2)

: general point

(1 + t, 2 + t, 3 + 2 t)

.

The second given line

\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}

can be rewritten with parameter

s

x = \frac{11 + 2 s}{3}, y = \frac{11 + s}{3}, z = \frac{19 + 2 s}{3}

s = 0

this gives the point

Q = (\frac{11}{3}, \frac{11}{3}, \frac{19}{3})

, and its direction ratios are

(2, 1, 2)

.

Find where this line meets L:

\frac{11 + 2 s}{3} = 1 + t, \frac{11 + s}{3} = 2 + t, \frac{19 + 2 s}{3} = 3 + 2 t

The first two give

2 s - 3 t = - 8

and

s - 3 t = - 5

. Subtracting:

s = - 3

, then

t = \frac{2}{3}

.

Foot point on L:

t = \frac{2}{3} \Rightarrow P = (\frac{5}{3}, \frac{8}{3}, \frac{13}{3})

.

Distance

QP

QP = (\frac{11}{3} - \frac{5}{3})^{2} + (\frac{11}{3} - \frac{8}{3})^{2} + (\frac{19}{3} - \frac{13}{3})^{2} = 2^{2} + 1^{2} + 2^{2} = 9 = 3

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option