Vector Algebra Question 137
Question: The equation of plane passing through a point $ A(2,-1,,3) $ and parallel to the vectors $ \mathbf{a}=(3,,0,-1) $ and $ \mathbf{b}=(-3,2,,2) $ is
[Orissa JEE 2005]
Options:
A) $ 2x-3y+6z-25=0 $
B) $ 2x-3y+6z+25=0 $
C) $ 3x-2y+6z-25=0 $
D) $ 3x-2y+6z+25=0 $
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Answer:
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 $ .
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