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Standard XII

Mathematics

Distance Formula

If the plane ...

Question

If the plane $x + 2 y - 2 z = d, d > 0$ is at a 5 unit distance from the point $(1, - 2, 1)$ ,then d = ____________.

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Solution

For given plane, x + 2y – 2z = d
Distance of this plane from (1, –2, 1) is 5 is given
$i . e . 5 = \frac{|(1) + 2 (- 2) - 2 (1) - d|}{\sqrt{1^{2} + 2^{2} + 2^{2}}} i . e . 5 = \frac{|1 - 4 - 2 - d|}{\sqrt{9}} i . e . 15 = |- 5 - d| i . e . 15 = 5 + d i . e . d = 10$

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If the plane x+2y-2z=d, d>0 is at a 5 unit distance from the point (1, -2, 1),then d = ____________.

For given plane, x + 2y – 2z = d Distance of this plane from (1, –2, 1) is 5 is given i.e. 5=1+2-2-21-d12+22+22i.e. 5=1-4-2-d9i.e. 15=-5-di.e. 15=5+di.e. d=10

If the plane $x + 2 y - 2 z = d, d > 0$ is at a 5 unit distance from the point $(1, - 2, 1)$ ,then d = ____________.

For given plane, x + 2y – 2z = d
Distance of this plane from (1, –2, 1) is 5 is given
$i . e . 5 = \frac{|(1) + 2 (- 2) - 2 (1) - d|}{\sqrt{1^{2} + 2^{2} + 2^{2}}} i . e . 5 = \frac{|1 - 4 - 2 - d|}{\sqrt{9}} i . e . 15 = |- 5 - d| i . e . 15 = 5 + d i . e . d = 10$