A) \[x-y+z=1\]
B) \[x+y+z=5\]
C) \[x+2y-z=1\]
D) \[2x-y+z=5\]
Correct Answer: A
Solution :
Key Idea: If a plane contains a line, then the normal to the plane is perpendicular to the line. Any plane passing through (3, 2, 0) is \[a(x-3)+b(y-2)+c(z-0)=0\] ... (i) It passes through\[(3,\,\,6,\,\,4)\] \[\therefore \] \[0.a+4b+4c=0\] ... (ii) Normal to plane (i) is perpendicular to given line \[\therefore \] \[a+5b+4c=0\] ... (ii) On solving Eqs. (i) and (ii), we get \[\frac{a}{1}=\frac{b}{-1}=\frac{c}{1}=k\] So, \[a=k,\,\,b=-k,\,\,c=k\] Putting the value of\[a,\,\,\,b,\,\,\,c\]in Eq. (i), we get \[x-y+z=1\] Note: In any line of a plane it has only one direction cosines and direction ratios may be more than one.
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