The angle between the lines \;x-2/3 = \;y+1/-2, z=2 and \;x-1/1 = \;y+3/3 = \;z+5/2 is

Correct Answer is : cos⁡−1( −3182)^{-1} ≤ft( \;{-3}{{182}} )cos−1(182−3)

Correct Answer is : cos⁡−1( −3182)^{-1} ≤ft( \;{-3}{{182}} )cos−1(182−3)

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CGPET 2011 Solved Paper

Section: Mathematics

Question : 118 of 150

Marks: +1, -0

\;\frac{x-2}{3} = \;\frac{y+1}{-2}

z=2

\;\frac{x-1}{1} = \;\frac{y+3}{3} = \;\frac{z+5}{2}

Solution:

Given lines are

\;\;\frac{x-2}{3} = \;\frac{y+1}{-2} = \;\frac{z-2}{0}

. . . (i)

\;\;\text{ and } \;\;\;\frac{x-1}{1} = \;\frac{y+3}{3} = \;\frac{z+5}{2}

. . . (ii)

\;\;\text{ Here, } \; a_1=3, b_1=-2, c_1=0

\;\;\text{ If } \; \theta \;\text{ be the angle between both the lines, then } \;

\; a_2=1, b_2=3, c_2=2

\;\cos \theta = \;\frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}

\;\Rightarrow \;\; \theta = \cos^{-1} \left( \;\frac{(3)(1)+(-2)(3)+(0)(2)}{\sqrt{9+4+0} \sqrt{1+9+4}} \right)

\;\Rightarrow \;\; \theta = \cos^{-1} \left( \;\frac{-3}{\sqrt{13} \sqrt{14}} \right)

\;\Rightarrow \;\; \theta = \cos^{-1} \left( \;\frac{-3}{\sqrt{182}} \right)

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