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Question

For the following planes, find the direction cosines of the normal to the plane and the distance of the plane from the origin.
3y+5=0.

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Solution

Given equation in vector form is r(0^i1^j+0^k)=53 [y=53].
Clearly, |0^i1^j+0^k|=02+(1)2+02=1
r^n=p, where ^n=(0^i1^j+0^k) and p=53.
Direction cosines of ^n are 0,1,0 and p=53.

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