A) \[x+5y-6z+19=0\]
B) \[x+5y-6z-19=0\]
C) \[x-5y-6z+19=0\]
D) \[x+5y+6z+19=0\]
Correct Answer: A
Solution :
[a] Given points be \[\left( 3,4,-1 \right)\] and\[\left( 2,-1,5 \right)\]. So, the direction ratio of the line joining of these points be \[\left( 3-2,4+1,-1-5 \right)=\left( 1,5,-6 \right)\] The equation of the plane perpendicular to this line, whose d.r. be\[\left( 1,5,-6 \right)\], be \[x+5y-6z=\lambda \] ...........(1) Since, the plane passes through the point \[\left( 2,-3,1 \right).\] \[\therefore 2+5(-3)-6\times 1=\lambda \] \[\Rightarrow 2-15-6=\lambda ~~~~~~~\Rightarrow \lambda =-19\] Hence, the required equation of the plane be \[x+5y-6z=-19\] \[\Rightarrow x+5y-6z+19=0\] Hence, option [a] is correct.You need to login to perform this action.
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