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.

Expert Solution

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.

=> 5x+2y+z=8.

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).

hkg10 answered 3 years ago

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