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Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

The distance ...

Question

The distance of the point (1,−2,4) from the plane passing through the point (1,2,2) and perpendicular to the planes $x - y + 2 z = 3$ and $2 x - 2 y + z + 12 = 0$ is :

2 \sqrt{2}

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4

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\sqrt{2}

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23

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Solution

The correct option is A $2 \sqrt{2}$

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Similar questions

Q. The distance of the point

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from the plane passing through the point

(1, 2, 2)

and perpendicular to the planes

x - y + 2 z = 3

and

2 x - 2 y + z + 12 = 0

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A plane passes through (1, -2, 1) and is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4,$ then the distance of the plane form the point (1, 2, 2) is

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The distance of the point (1,−2,4) from the plane passing through the point (1,2,2) and perpendicular to the planes x−y+2z=3 and 2x−2y+z+12=0 is :

The correct option is A 2√2

The distance of the point (1,−2,4) from the plane passing through the point (1,2,2) and perpendicular to the planes $x - y + 2 z = 3$ and $2 x - 2 y + z + 12 = 0$ is :

The correct option is A $2 \sqrt{2}$