In: Advanced Math
Find the equation of the plane through the point (1,1,1) which is perpendicular to the line of intersection of the two planes x−y−3z=−1 and x−3y+z= 2.
The direction of intersection line L is v.
n1 is perpendicular to v, and n2 is perpendicular to v,
so v = n1 X n2 = (-10,-4,-2)
The plane we want is perpendicular to L, so v is also the normal vector of the plane. Thus, its equation is -10(x-1)-4(y-1)-2(z-1)=0.
The normal vector n1 of x−y−3z=−1 is n1=(1,-1,-3).
The normal vector n2 of x−3y+z= 2 is n2=(1,-3,1).