Find the equation of the plane passing through the point (−1,3,2) and perpendicular to each of the planes x+2y+3z=5 and 3x+3y+z=0.
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Solution
The equation of the plane passing through (−1,2,3) will be given by a(x+1)+b(y−2)+c(z−3)=0, where a,b and c are the direction ratios of the normal to the plane.
Also, this plane is perpendicular to the planes x+2y+3z=5 and 3x+3y+z=0. According to the perpendicularity condition of planes, the dot product of the direction ratios of the normals to the two planes should be 0.
⇒a+2b+3c=0 and 3a+3b+c=0
⇒c=−a+2b3
Substituting this in the second equation,
3a+3b−a+2b3=0
⇒9a+9b−a−2b=0
⇒8a+7b=0
⇒b=−8a7
⇒c=−a+2b3
=−a−16a73
=9a7×3
=3a7
Substituting these values of b and c in the equation of the plane,
a(x+1)−8a7(y−2)+3a7(z−3)=0
⇒7(x+1)−8(y−2)+3(z−3)=0
⇒7x−8y+3z+7+16−9=0
⇒7x−8y+3z+14=0
Hence, the equation of the required plane is 7x−8y+3z+14=0