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Question
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
Solution
5y + 8 = 0
⇒ 0x − 5y + 0z = 8 … (1)
The direction ratios of normal are 0, −5, and 0.
`:. sqrt(0+(-5)^2 + 0)` = 5
Dividing both sides of equation (1) by 5, we obtain
`-y = 8/5`
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 0, −1, and 0 and the distance of normal from the origin is `8/5` units.
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