Q.110. Equation of a Plane.

Question 110.
Find an equation of the plane that passes through the line of intersection of the planes
x - z = 2 and y + 4z = 1 and is perpendicular to the plane x + y - 2z = 3.

Answer 110.
A plane passing through the line of intersection of the planes
x - z = 2 and y + 4z = 1 has equation of the form
(x - z - 2) + k*(y + 4z - 1) = 0
=> x + ky + (4k - 1)z - k - 2 = 0 ... (1)
Its normal is
n1 = [1, k, (4k-1)]

It is perpendicular to the plane
x + y - 2z = 3 whose normal is
n2 = (1, 1, -2)

As these two planes are perpendicular to each other,
n1 . n2 = 0
=> [1, k, (4k-1)] . (1, 1, -2) = 0
=> 1 + k - 2(4k-1) = 0
=> -7k = - 3
=> k = 3/7

Plugging this value of k in ( 1 ),
x + (3/7)y + (12/7 - 1)z - 3/7 - 2 = 0
=> 7x + 3y + 5z = 17.

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