Three Dimensional Geometry Question 401
Question: Let A (1, 1, 1), B (2, 3, 5) and C ( $ - $ 1,0, 2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is
Options:
A) $ 2x-3y+z+2\sqrt{14}=0 $
B) $ 2x-3y+z-\sqrt{14}=0 $
C) $ 2x-3y+z+2=0 $
D) $ 2x-3y+z-2=0 $
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Answer:
Correct Answer: A
Solution:
[a] A (1, 1, 1), B(2, 3, 5), C (-1, 0, 2) direction ratios of AB are <1, 2, 4>. Direction ratios of AC are $ <-,2,-1,1>. $ Therefore, directions ratios of normal to plane ABC are $ <2,,-3,1> $ As a result, equation of the plane ABC is $ 2x-3y+z=0 $ . Let the equation of the required plane be $ 2x-3y+z=k. $ Then $ | \frac{k}{\sqrt{4+9+1}} |=2 $ Or $ k=\pm 2\sqrt{14} $ Hence, equation of the required plane is $ 2x-3y+z+2\sqrt{14}=0 $
BETA
