Q.362. To find eqn. of a plane containing a line and perpendicular to a given plane.

Question 362.
Find the equation of the plane containing the line: r = (1,2,1) + k (3,1,0)
and perpendicular to the plane 3x+4y-2z = 13.

Answer 362.
Normal, n, of the required plane is perpendicular to the direction (3, 1, 0) of the line
and the normal (3, 4, -2) of the given plane

=> n = (3, 1, 0) x (3, 4, -2)
= (-2, 6, 9)
Also the plane contains (1, 2, 1) which is a point on the given line
=> eqn. of the plane is of the form
-2x + 6y + 9z = k
The point (1, 2, 1) of the line is on this plane
=> k = 19
=> eqn. of the required plane is
-2x + 6y + 9z = 19
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Verification:
If any two points of the line lies on this plane, then the line is in the plane.
(1, 2, 1) is a point on the line that lies on the plane.
k = 1 => (4, 3, 1) is another point of the line.
Plugging in the eqn. of the plane, it is satisfied.
=> given line is on the plane found.
Direction of the normals of the given and found planes are (3, 4, -2) and (-2, 6, 9)
(3, 4, -2) . (-2, 6, 9)
= -6 + 24 - 18
= 0
=> the plane found is perpendicular to the given plane.

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