Test Index

KEAM 2014 Math Paper

Question : 10 of 120

Marks: +1, -0

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

x=3r+2; y=-2r-1; z=2,

where r is a parameter, is

Solution:

Given equation of lines are,

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

and

x=3r+2 ; y=-2r-1 ; z=2

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

and

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

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

and

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

DR's of lines Ist and IInd lines are

\left(1, \;\frac{3}{2}, 2\right)

and

(3,-2,0)

\therefore

The angle between two straight lines is

\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}}

= \;\frac{1 \times 3 - \frac{3}{2} \times 2 + 2 \times 0}{\sqrt{1^{2}+\left(\;\frac{3}{2}\right)^{2}+2^{2} \sqrt{3^{2}+2^{2}+0^{2}}}}

= \;\frac{3-3+0}{\sqrt{1+\;\frac{9}{4}+4} \sqrt{9+4}}

= \;\frac{0}{\sqrt{\;\frac{29}{4}} \sqrt{13}}

\Rightarrow \;\; \cos \theta = 0

\Rightarrow \;\; \theta = \frac{\pi}{2}

Go to Question:

Prev Question

Next Question