Write the equation of the plane passing through the point 1

Write the equation of the plane passing through the point (1, 2, 3) parallel to the plane 2x - y + 3z = 0 Write the equation of the plane that passes through the point (1, 2, 3) and is perpendicular to the line of intersection of the planes x - 2y + 3r = 0 and 2x + 7y - z = 0

Solution

(a) Equation of plane parallel to the given plane : 2x - y + 3z = k where k is any real number

Plugging the given point in the above equation

=> 2(1) - (2) + 3(3) = k

=> 9 = k

=> Required equation of plane : 2x - y + 3z = 9

(b) Normal vector to the plane x - 2y + 3z = 0 => < 1 , -2 , 3 >

Normal vector to the plane 2x + 7y - z = 0 => < 2 , 7 , -1 >

The vector which would be perpendicular to both the normal vectors would be given by cross product of these vector

< 1 , -2 , 3 > x < 2 , 7 , -1 > = < -19 , 7 , 11 >

Therefore , the equation of plane perpendicular to the two planes : -19x + 7y + 11z = k where k is any real number

Plugging the given point we get

=> -19(1) + 7(2) + 11(3) = k

=> k = 28

The required plane : -19x + 7y + 11z = 28