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KEAM 2013 Math Paper

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Question : 75 of 120

Marks: +1, -0

If the lines

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

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

are perpendicular to each other, then p is equal to

1
$-1$
10
$-\frac{7}{5}$
$-19$

Solution:

Given lines are

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

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

.....(i)

and

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

....(ii)

since, both lines are perpendicular.

\therefore\;\; (1)(2)+(-1)(5)+(3)(p)=0

(by perpendicularity condition)

\Rightarrow\;\; 2-5+3p=0 \Rightarrow 3p=3

\therefore\;\; p=1

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