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Question : 31 of 160
Marks:
+1,
-0
Solution:
The equation of the given line is
===r The point on the line is,
P(2r+1,r+2,2r−3) The equation of plane is,
(r−(++)).[{(++) −((−−))}×(−2)]=0 (r−(++)).[(2+2) ×(−2)]=0 (r−(++)) .[(4+2−2)]=0 4x+2y−2z=4 Therefore,
4(2r+1)+2(r+2)−2(2r−3)=4 r=− Hence, the point
P is
P(−,,) =(−7+−19)
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