If the lines (x-1)/2 = (y+1)/3 = (z-1)/4 and (x-3)/1 = (y-k)/2 = z/1 intersect, then the value of k is

If the lines (x-1)/2 = (y+1)/3 = (z-1)/4 and (x-3)/1 =… — JEE Main 2004 Mathematics PYQ

AIEEE 2004

If the lines

\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}

and

\dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1}

intersect, then the value of

k

A.
$\dfrac{3}{2}$
B.
$\dfrac{9}{2}$
C.
$-\dfrac{2}{9}$
D.
$-\dfrac{3}{2}$

Show answer and solution

Correct option: B

Solution

For two lines to intersect, the vectors connecting a point on each line and their direction vectors must be coplanar.

Point on L1: A = (1, -1, 1), direction d1 = (2, 3, 4)
Point on L2: B = (3, k, 0), direction d2 = (1, 2, 1)

Vector AB = (3-1, k-(-1), 0-1) = (2, k+1, -1)

Coplanarity condition: AB · (d1 × d2) = 0

d1 × d2 = |i j k; 2 3 4; 1 2 1| = i(3·1 - 4·2) - j(2·1 - 4·1) + k(2·2 - 3·1)
= i(3-8) - j(2-4) + k(4-3) = (-5, 2, 1)

AB · (-5, 2, 1) = 2(-5) + (k+1)(2) + (-1)(1) = -10 + 2k + 2 - 1 = 2k - 9 = 0

So k = 9/2.

[AIEEE 2004]

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