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Question
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
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Solution
The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1)
The direction ratios of normal are 0, 0, and 1.
`:.sqrt(0^2 + 0^2 + 1^2) = 1`
Dividing both sides of equation (1) by 1, we obtain
0.x + 0.y + 1.z = 2
This is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.
Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.
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