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# Find the equation of the plane passing through the point (-1, - 1, 2) and perpendicular to each of the following planes:2x + 3y - 3z = 2 and 5x - 4y + z = 6

Asked by Topperlearning User 2nd March 2015, 2:30 PM

The equation of the plane passing through the point (-1, -1, 2) is:

a(x + 1) + b(y + 1) + c (z - 2) = 0...(1)

Where a, b c are the direction ratios of the normal to the plane

It is given that the plane (1) is perpendicular to the planes.

2x +3y - 3z = 2 and 5x - 4y + z = 6

2a + 3b - 3c = 0...(2)

5a - 4b + c = 0...(3)

Solving equations (2) and (3), we have:

So the direction ratios of the normal to the required plane are multiples of 9, 17, and 23.

Thus, the equation of the required plane is:

Answered by Expert 2nd March 2015, 4:30 PM
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