Three Dimensional Geometry Question 126
Question: The direction ratios of the line $ x-y+z-5= $ $ 0=x-3y-6 $ are
[MP PET 1999; Pb. CET 2000]
Options:
A) 3, 1, ? 2
B) 2, ? 4, 1
C) $ \frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}} $
D) $ \frac{2}{\sqrt{41}},\frac{-4}{\sqrt{41}},\frac{1}{\sqrt{41}} $
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Answer:
Correct Answer: A
Solution:
If l, m, n are direction ratios of line, then by $ Al+Bm+Cn=0 $ For $ x-y+z-5=0,l-m+n=0 $ ?..(i) For $ x-3y-6=0,l-3m+0n=0 $ ?..(ii) or $ \frac{l}{0+3}=\frac{m}{1-0}=\frac{n}{-3+1} $ or $ \frac{l}{3}=\frac{m}{1}=\frac{n}{-2} $
$ \therefore , $ Direction ratios are $ (3,1,-2) $ . Note : Option , $ ( \frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}} ) $ may also be an answer but best answer is $ A(3,1,-2) $ because in direction cosines are written.
BETA
