The lines `(x-1)/(2)=(y+1)/(2)=(z-1)/(4) and (x-3)/(1)=(y-k)/(2)=(z)/(1)` intersect each other at point

Options

A. `(-2,-4,5)`
B. `(-2,-4,-5)`
C. `(2,-4,-5)`
D. `(2,-4,-5)`

Answer

b

Explanation

Given lines are
`(x-1)/(2) = (y+1)/(2) = (z-1)/(4) = lambda` (say) `"……."(i)`
and `(x-3)/(1) = (y-6)/(2) = z/1 "……."(ii)`
Any point on the line (i) is
`P(2lambda + 1, 2lambda - 1, 4 lambda + 1)`
Put `x = 2 lambda + 1, y = 2 lambda - 1`
and `z = 4 lambda + 1`
in Eq. (i) we get
`(2lambda + 1 - 3)/(1) = (2lambda - 1- 6)/(2) = (4lambda + 1)/(1)`
Taking first two terms, we get
`4lambda - 4 = 2lambda - 7`
`rArr 2 lambda = -3`
Put `lambda = - 3/2` in Eq . (ii), we get
Point of intersection is
`P [(2 xx -3/2 + 1,2 xx(-3/2) - 1), 4 xx(-3/2) + 1]`
` -=P (-2,-4,-5)`