the distance of the point `(2,3,4)` from the line `(1-x)=y/2=1/3(1+z)`

Options

A. `1/7 sqrt35`
B. `4/7 sqrt35`
C. `2/7 sqrt35`
D. `3/7 sqrt35`

Answer

D

Explanation

Given `(x-1)/-1=(y-0)/(2)=(z+1)/(3)=r......(i)`
Then, coordinates of any point N on the line (i) are
`(-r+1, 2r, 3r-1).......(ii)`
Let N be the foot of the perpendicular to line (i),
`therefore` Direction ratios of PN are
`(-r+1-2, 2r-3,3r-1-4)=(-r-1,2r-3,3r-5).....(iii)`
PN is perpendicular to line (i),
Using the condition, `a_1a_(2)+b_1b_2+c_1c_2=0`
`Rightarrow -1(-r-1)+2(2r-3)+3(3r-5)=0`
`Rightarrow r+1+4r-6+9r-15=0`
`Rightarrow r=10/7`
Then from Eq. (ii) we get
`N=(-10/7+1 , 20/7, 30/7-1)=(-3/7,20/7,23/7)`
Now perpendicular distance `PN=sqrt((-3/7-2)^(2)+(20/7-3)^(2)+(23/7-4)^2)`
`=sqrt(((-17)/(7))^(2)+((-1)/(7))^(2)+((-5)/(7))^(2)`
`1/7 sqrt(289+1+25)=3/7 sqrt35`