The equation of a plane through the point `(2, 3, 1)` and `(4, -5, 3)` and parallel to x-axis

Options

A. `x+y-3z=2`
B. `y+4z=7`
C. `y+3z=6`
D. `x+5y-3z=4`

Answer

b

Explanation

Let the required equation be `by+cz+d=0`………….(i)
Since it passe through the points A(2,3,1) and B(4,-5,3), we have
`3b+c+d=0`………..(ii)
`-5b+3c+d=0`………(iii)
On solving (ii) and (iii) by cross multiplication, we have
`b/(1-3)=c/(-5-3)=d/(9+5)`
`rArr vecr.vecn=5sqrt(14)`.
`rArr b/1=c/4=d/-7=k` (say).
Then, b=k,c=4k, and d=-7k.
Substituting these values in (i), we get the required equation ias
`ky+4kz-7k=0 rArr y+4z-7=0 rArr y+4z=7`.