Convert plane 2x+3y+4z=5 into a perpendicular and find its cosines.
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Step-by-step explanation:
Given equation of the plane
2x+3y+4z=5
divide with √a square+ b square+ c square
√1+4+4=3 on both sides
2/3x+3/3y+4/3z=5/3
2/3x+y+4/3z=5/3
which is of the form lx+my+nz=p
where (l,m,n) are D.C's of the plane
DC's of normal plane(l,m,n)=(2/3,1,4/3)
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