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Mathematics Question on Three Dimensional Geometry

The two lines x=my+n,z=py+qx = my + n, z = py + q and x=m′y+n′, z=p′y+q′x = m'y + n',\, z = p'y + q' are perpendicular to each other, if

A

mm′+pp′=1mm' + pp' = 1

B

mm′+pp′=−1\frac{m}{m'} + \frac{p}{p'} = -1

C

mm′+pp′=1\frac{m}{m'} + \frac{p}{p'} = 1

D

mm′+pp′=−1mm' + pp' = -1

Answer

mm′+pp′=−1mm' + pp' = -1

Explanation

Solution

Given lines are
x=my+n,z=py+qx = my + n, z = py + q
and x=m′ y+n′, z=p′ y+q′x = m'\, y + n',\, z = p'\, y + q'
Above equations can be rewritten as
x−nm=y−01=z−qp\frac{x-n}{m} = \frac{y-0}{1} = \frac{z-q}{p}
and x−n′m′=y−01=z−q′p′\frac{x-n'}{m'} = \frac{y-0}{1} = \frac{z-q'}{p'}
Lines will be perpendicular, if
mm′+1+pp′=0mm' + 1 + pp' = 0
⇒mm′+pp′=−1\Rightarrow mm' + pp' = -1