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Question
The foot of the perpendicular drawn from the (- 1, - 3, - 5) to a plane is (2, 4, 6). The equation of the plane is:
The foot of the perpendicular drawn from the (- 1, - 3, - 5) to a plane is (2, 4, 6). The equation of the plane is:
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Solution
The correct option is C 3x + 7y + 11z = 100
Since, the foot of the perpendicular to the plane is P(2, 4, 6). Therefore, (2, 4, 6) is the point on the plane.
So, equation of the plane passing through the point (2,.4, 6) is:
a(x - 2) + b(y - 4) + c(z - 6) = 0.
Now, the direction ratios of the perpendicular line AP are 3, 7, 11.
Therefore, the required plane is:
3(x - 2) + 7(y - 4) + 11(z - 6) = 0
i.e, 3x + 7y + 11z = 100.
3x + 7y + 11z = 100
Since, the foot of the perpendicular to the plane is P(2, 4, 6). Therefore, (2, 4, 6) is the point on the plane.
So, equation of the plane passing through the point (2,.4, 6) is:
a(x - 2) + b(y - 4) + c(z - 6) = 0.
Now, the direction ratios of the perpendicular line AP are 3, 7, 11.
Therefore, the required plane is:
3(x - 2) + 7(y - 4) + 11(z - 6) = 0
i.e, 3x + 7y + 11z = 100.
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