Equation of the plane containing the straight line `x/2=y/3=z/4` and perpendicular to the plane containing the straight lines `x/2=y/4=z/2` and `x/4=y/2=z/3` is

Options

A. `5x+2y-4z=0`
B. `x+2y-2z=0`
C. `ax+2y-3z=0`
D. `x-2y+z=0`

Answer

D

Explanation

Let `P_(1)` be the plane containing the lines
`(x)/(3)=(y)/(4)=(z)/(2)" and "(x)/(4)=(y)/(2)=(z)/(3).`
For these two lies, direction vectors are
`b_(1)=3hati+4hatj+2hatk" and "b_(2)=4hati+2hatj+3hatk.`
A vector along the normal to the plane `P_(1)` is given by
`n_(1)=b_(1)xxb_(2)=|{:(hati,hatj,hatk),(3,4,2),(4,2,3):}|`
`hati(12-4)-hatj(9-8)+hatk(6-16)=8hati-hatj-10hatk`
Let `P_(2)` be the plane containing the line `(x)/(2)=(y)/(3)=(z)/(4)` and perpendicular to plane` P_(1).`
For the line `(x)/(2)=(y)/(3)=(z)/(4)`, the direction vector is
`b=2hati+3hatj+4hatk` and it passes through the point with position vector `a=0hati+0hatj+0hatk.`
`becauseP_(2)` is perpendicular to `P_(1)`, therefore `n_(1)` and b lies along the plane.
Also, `P_(2)` also passes through the point with position vector a.
`:.` Equation of plane `P_(2)` is given by
`(r-a).(n_(1)xxb)=0implies|{:(x-0,y-0,z-0),(8,-1,-10),(2," "3," "4):}|=0`
`impliesx(-4+30)-y(32+20)+z(24-2)=0`
`implies26x-52y+26z=0`
`implies" "x-2y+z=0`