A) \[2x+3y+z=16\]
B) \[2x+3y-z=14\]
C) \[2x+3y+z=14\]
D) \[2x+3y-z=0\]
Correct Answer: B
Solution :
[b] Since, coordinates of points O and P are (0, 0, 0) and (2, 3, -1), respectively. Direction ratios of OP are <2, 3, -1>. The plane is perpendicular to OP, so, its equation is \[2x+3y-z+d=0\] (i) Since, this plane passes through \[(2,3-1);2\times 2+3\times 3-1\times -1+d=0\] \[\Rightarrow 4+9+1+d=0\] \[\Rightarrow d=-14\] On putting the value of d in equation (i) \[2x+3y-z-14=0\] \[\Rightarrow 2x+3y-z=14\] Which is required equation of plane.
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