A) \[2x-3y+6z-25=0\]
B) \[2x-3y+6z+25=0\]
C) \[3x-2y+6z-25=0\]
D) \[3x-2y+6z+25=0\]
Correct Answer: A
Solution :
As plane is parallel to a given vector Þ Normal of plane must perpendicular to the given vectors. Given point to which plane passes through is (2, ?1,3). Let A, B, C are direction ratios of its normal. \ Equation of plane is, \[A(x-2)+B(y+1)+C(z-3)=0\] ?..(i) Now normal to plane \[A\mathbf{i}+B\mathbf{j}+C\mathbf{k}\] is perpendicular to the given vectors \[\mathbf{a}=3\mathbf{i}+0\mathbf{j}-\mathbf{k}\] and \[\mathbf{b}=-3\mathbf{i}+2\mathbf{j}+2\mathbf{k}\] \ \[3A+0B-C=0\] ?..(i) \[-3A+2B+2C=0\] .....(ii) Solving (i) and (ii) we get, \[\frac{A}{2}=\frac{B}{-3}=\frac{C}{6}\] \Equation of plane be \[2(x-2)-3(y+1)+6(z-3)=0\] i.e., \[2x-3y+6z-25=0\].You need to login to perform this action.
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