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Question
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3y – z = 5
Solution
2x + 3y − z = 5 … (1)
The direction ratios of normal are 2, 3, and −1.
`:.sqrt((2)^2 + (3)^3 + (-1)^2) = sqrt(14)`
Dividing both sides of equation (1) by `sqrt14`, we obtain
This equation 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 normal from the origin.
Therefore, the direction cosines of the normal to the plane are `2/sqrt14, 3/sqrt14 and (-1)/sqrt(14)` the distance of normal from the origin is `5/sqrt14` units.
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