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

Options

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

Answer

C

Explanation

The equation of a plane containing the straight line `x/2=y/3=z/4` is
`ax+by+cz=0`………….i
where `2a+3b+4c=0`………….ii
Let the direction ratios of the normal to the plane containing the straight lines `x/3=y/4=z/2` and `x/4=y/2=z/3` be proportional to `alpha, beta, gamma`. Then
`3alpha+4beta +2gamma=0`
and `4alpha+2beta+3gamma=0`
`:. (alpha)/8=(beta)/(-1)=(gamma)/(-10)`
It is given that the plane (i) is perpendicular to the plane the direction ratios of normal to which are proportional to 8,-1,-10.
`:.8a-b-10c=0`.............iii
From i and iii we have
`a/(-30+4)=b/(32+20)=c/(-2-24)`
`impliesa/(-26)=b/52-c/(-26)`
`implies a/a=b/(-2)=c/1`
Substituting the values of `a,b,c ` in (ii) we obtain `x-2y+z=0` as the required plane.
ALITER As the line `x/2=y/3=z/4` passes through the origin. So plane containing it also passes through the origin. Let `vecn` be a vector normal to the required plane. Then `vecn` is perpendicular to `veca=2hati+3hatj+4hatk` an `vecbxxvecc` where `vecb=3hati+4hatj+2hatk` and `vecc=4hati+2hatj+3hatk`.
`:.vecn=vecaxx(vecbxxvecc)`
`impliesvecn=(veca.vecc)vecb-(veca.vecb)vecc`
`impliesvecn=26(3hati+4hatj+hatk)-26(4hati+2hatj+3hatk)=26(-hati+hatj-hatk)`
Hence the equation of the plane is `vecr.vecn=0`
`impliesvecr.(-hati+2hatj-hatk)=0` or `x-2y+z=0`