Three Dimensional Geometry Question 236
Question: The acute angle between the line joining the points (2,1,?3), (?3,1,7) and a line parallel to $ \frac{x-1}{3}= $ $ \frac{y}{4}=\frac{z+3}{5} $ through the point (?1, 0, 4) is
[MP PET 1998]
Options:
A) $ {{\cos }^{-1}}( \frac{7}{5\sqrt{10}} ) $
B) $ {{\cos }^{-1}}( \frac{1}{\sqrt{10}} ) $
C) $ {{\cos }^{-1}}( \frac{3}{5\sqrt{10}} ) $
D) $ {{\cos }^{-1}}( \frac{1}{5\sqrt{10}} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
Direction ratio of the line joining the point $ (2,1,-3),, $ $ ,(-,3,1,7) $ are $ (a_1,b_1,c_1), $ $ ,\Rightarrow (-,3-2,1-1,7-(-3))\Rightarrow (-,5,0,10) $ Direction ratio of the line parallel to line $ \frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5} $ are $ (a_2,,b_2,c_2)\Rightarrow (3,4,5) $ Angle between two lines, $ \cos \theta =\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2}}\sqrt{a_2^{2}+b_2^{2}+c_2^{2}}} $ $ \cos \theta =\frac{(-,5\times 3)+(0\times 4)+(10\times 5)}{\sqrt{25+0+100}\sqrt{9+16+25}} $ $ \cos \theta =\frac{35}{25\sqrt{10}}\Rightarrow \theta ={{\cos }^{-1}}( \frac{7}{5\sqrt{10}} ) $ .
BETA
