JEE Main On 16 April 2018 Question 15
Question: If the angle between the lines, $ \frac{x}{2}=\frac{y}{2}=\frac{z}{1} $ and $ \frac{5-x}{-2}=\frac{7y-14}{P}=\frac{z-3}{4} $ is $ {{\cos }^{-1}}( \frac{2}{3} ) $ , then p is equal to? [JEE Main 16-4-2018]
Options:
A) $ -\frac{7}{4} $
B) $ \frac{2}{7} $
C) $ -\frac{4}{7} $
D) $ \frac{7}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
Above formula is used to find angle $ \theta $ between two lines having direction ratios $ a _1,b _1,c _1 $ and $ a _2,b _2,c _2 $
Now let us find the direction cosines of the given lines
$ \frac{x}{2}=\frac{y}{2}=\frac{z}{1} $ direction cosines are 2, 2, 1 next line can also be written as
$ \frac{x-5}{2}=\frac{y-2}{\frac{P}{7}}=\frac{z-3}{4} $ Hence direction cosines are $ 2,\frac{P}{7},4 $
and we know that $ \cos \theta =\frac{2}{3} $ Keeping the value of these cosines in above formula we get
$ \frac{2}{3}=| \frac{2\times 2+2\times \frac{P}{7}+1\times 4}{\sqrt{2^{2}+2^{2}+1^{2}}\sqrt{2^{2}+\frac{P^{2}}{49}+4^{2}}} |=\frac{8+\frac{2P}{7}+4}{3\times \sqrt{2^{2}+\frac{P^{2}}{49}+4^{2}}} $ After cross multiplication we get
$ \Rightarrow $ $ {{( 4+\frac{P}{7} )}^{2}}=20+\frac{P^{2}}{49} $
$ \Rightarrow $ $ 16+\frac{8P}{7}+\frac{P^{2}}{49}=20+\frac{P^{2}}{49} $
$ \Rightarrow $ $ \frac{8P}{7}=4 $
$ \Rightarrow $ $ P=\frac{7}{2} $
Therefore Answer is D $ \cos | \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}}} | $
