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A plane passes through (1,-2,1) and is ...

A plane passes through (1,-2,1) and is perpendicualr to two planes` 2x-2y+z=0" and "x-y+2z=4,` then the distance of the plane from the point (1,2,2) is

A

0

B

1

C

`sqrt2`

D

`2sqrt2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the equation of the plane that passes through the point (1, -2, 1) and is perpendicular to the two given planes. Then we will calculate the distance from this plane to the point (1, 2, 2). ### Step 1: Identify the normal vectors of the given planes The equations of the two planes are: 1. \(2x - 2y + z = 0\) 2. \(x - y + 2z = 4\) The normal vector of the first plane \(2x - 2y + z = 0\) is \(\mathbf{n_1} = (2, -2, 1)\). The normal vector of the second plane \(x - y + 2z = 4\) is \(\mathbf{n_2} = (1, -1, 2)\).
To solve the problem, we need to find the equation of the plane that passes through the point (1, -2, 1) and is perpendicular to the two given planes. Then we will calculate the distance from this plane to the point (1, 2, 2). ### Step 1: Identify the normal vectors of the given planes The equations of the two planes are: 1. \(2x - 2y + z = 0\) 2. \(x - y + 2z = 4\) ...
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