Three Dimensional Geometry Question 241
Question: The points $ A(-1,3,0) $ , $ B,(2,,2,,1) $ and $ C,(1,,1,,3) $ determine a plane. The distance from the plane to the point $ D(5,,7,8) $ is
[AMU 2001]
Options:
A) $ \sqrt{66} $
B) $ \sqrt{71} $
C) $ \sqrt{73} $
D) $ \sqrt{76} $
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Answer:
Correct Answer: A
Solution:
Equation of plane passing through $ (-1,,3,,0) $ is $ A(x+1)+B(y-3)+C(z-0)=0 $ ……(i) Also, plane (i) is passing through the points $ (2,,2,,1) $ and (1, 1, 3). So, $ 3A-B+C=0 $ …..(ii) $ 2A-2B+3C=0 $ …..(iii) Solving (ii) and (iii), $ \frac{A}{-3+2}=\frac{B}{2-9}=\frac{C}{-6+2} $ \ $ A:B:C=-1:-7:-4 $ or $ A:B:C=1:7:4 $ From (i), $ 1(x+1)+7(y-3)+4(z)=0 $ or $ x+7y+4z-20=0 $
$ \therefore $ Distance from the plane to the point (5, 7, 8) is, $ \frac{1\times 5+7\times 7+4\times 8-20}{\sqrt{1^{2}+7^{2}+4^{2}}}=\frac{5+49+32-20}{\sqrt{66}}=\frac{66}{\sqrt{66}}=\sqrt{66} $ .
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