A) \[2x+6y+3z=7\]
B) \[2x-6y+3z=7\]
C) \[2x+6y-3z=49\]
D) \[2x+6y+3z=49\]
Correct Answer: D
Solution :
Distance of point P from origin \[OP=\sqrt{4+36+9}=7\] Now d.r?s of OP = 2?0, 6 ? 0, 3 ? 0 = 2, 6, 3 \ d.c?s of OP = \[\frac{2}{7},\,\frac{6}{7},\,\frac{3}{7}\] \Equation of plane in normal form is \[lx+my+nz=p\] Þ \[\frac{2}{7}x+\frac{6}{7}y+\frac{3}{7}z=7\] Þ \[2x+6y+3z=49\].You need to login to perform this action.
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