Question

Find the Cartesian equation of the plane passing through the points `A (0,0,0)` and `b(3,-1,2)` and parallel to the line `(x-4)/1=(y+3)/(-4)=(z+1)/7`

Answer 1

Vector joining point `A` and `B` wil be,
`veca = 3hati-hatj+2hatk`
Given line is,
`(x-4)/1 = (y+3)/(-4) = (z+1)/7`
So, `vecb = hati-4hatj+7hatk` is the another vector in the plane.
So, a normal to the plane will be,
`vec a xx vec b`.
Now, `vec a xx vec b = |[hati,hatj,hatk],[3,-1,2],[1,-4,7]|`
`=hati(-7+8)-hatj(21-2)+hatk(-12+1)`
`=hati-19hatj-11hatk`
So, equation of plane will be,
`vecr *(vec a xx vecb) = 0`
Let `vecr = xhati+yhatj+zhatk`
So, required equation will be,
`(xhati+yhatj+zhatk)*(hati-19hatj-11hatk) = 0`
`=>x-19y-11z = 0`, which is the equation of the plane.