Solution:
Consider the points.
A=(2,3,−1)
B=(3,5,−3)
The DR's of AB are,
DR′sAB=(3−2,5−3,−3+1)
=(1,2,−2)
And
C=(1,2,3)
D=(3,y,7)
DR's CD =(3−1,y−2,7−3)
=(2,y−2,4)
It is given that AB⟂CD
Hence,
a1a2+b1b2+c1c2=0
(1)(2)+2(y−2)+(−2)(4)=0
2y−10=0
y=5
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