A) \[x-2=0\]
B) \[y+3=0\]
C) \[z-5=0\]
D) \[2x-3y+5z+4=0\]
Correct Answer: C
Solution :
Since, the plane perpendicular to z-axis, so the normal of the plane is parallel to z-axis, the direction cosine?s of z-axis is \[(0,0,1)\]. Then, the equation of plane which passes through the point \[(3,-3,5)\] and perpendicular to z-axis is \[0(x-2)+0(y+3)+1(z-5)=0\] \[\Rightarrow \] \[z-5=0\]You need to login to perform this action.
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