A) \[x-y+z=1\]
B) \[x+y+z=5\]
C) \[x+2y-z=1\]
D) \[2x-y+z=5\]
Correct Answer: A
Solution :
Any plane passing through (3, 2, 0) \[a(x-3)+b(y-2)+c(z-0)=0\] ?.(i) It passes through (4, 7, 4) Normal to plane (i) is perpendicular to given line \[a+5b+4c=0\] ...(ii) \[\frac{a}{1}=\frac{b}{-1}=\frac{c}{1}=k\] so \[a=k,b=-k,c=k\] Putting the value of a, b, c in Eq. (i) we get, \[x-y+z=1\]You need to login to perform this action.
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