A) \[x+2y+4z=12\]
B) \[4x+2y+z=12\]
C) \[x+2y+4z=3\]
D) \[4x+2y+z=3\]
E) \[x+y+z=12\]
Correct Answer: B
Solution :
Given, plane meets the co-ordinate axes at \[A(a,0,0),\,\,B(0,b,0)\,\,C(0,0,c)\] \ Centroid \[\equiv \left( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right)=(1,\,2,\,4)\] Þ \[a=3,b=6,c=12\] Hence, equation of required plane is, \[\frac{x}{3}+\frac{y}{6}+\frac{z}{12}=1\] Þ \[4x+2y+z=12\].
You need to login to perform this action.
You will be redirected in
3 sec
