We have to find the Cartesian equation of the plane that is perpendicular to the plane x-2y+z=6 and contains the line (x,y,z) = (2,-1,-1) + t(3,1,2).

The normal to the given plane, (1, -2, 1) lies in the plane to be determined. Another vector that lies in the plane to...

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We have to find the Cartesian equation of the plane that is perpendicular to the plane x-2y+z=6 and contains the line (x,y,z) = (2,-1,-1) + t(3,1,2).

The normal to the given plane, (1, -2, 1) lies in the plane to be determined. Another vector that lies in the plane to be determined is (3, 1, 2)

The cross product of the two is

`(1, -2, 1)ox(3, 1, 2)` = (-5, 1, 7)

The normal to the required plane is (-5, 1, 7)

If we equate t = 1, we get a point on the plane as (5, 0, 1)

Any vector (x - 5, y , z - 1) is perpendicular to (-5, 1, 7). The dot product of perpendicular vectors is 0.

=> -5(x - 5) + y + 7(z - 1) = 0

=> -5x + 25 + y + 7z - 7 = 0

=> 5x - y - 7z = 18

**The required plane is 5x - y - 7z = 18**