Question

find the equation of the plane passing through the point A=(3,-2,-1) and parallel to the vector b=i-2j+4k and c=3i+2j-5k​

Answer 1

equation of plane

Step-by-step explanation:

As equation of plane passing through the single point is

A(x-$x_{1}$) +B(y-$y_{1}$)+C(z-$z_{1}$) =0.

Therefore equation of plane will be A(x-3)+B(y+2)+C(z+1) =0 .

Here A,B,C are the direction ratios of normal to the plane. now this plane is parallel to the vector b and c given .

Therefore A.1+B(-2)+4C=0 and 3A+2B+4C=0. solving these two equations we get A=C/4 and B=17C/4 .

Now replacing this A and B in the starting equation we get 2x+17y+8z+36=0