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KEAM 2016 Math Paper

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Question : 76 of 120

Marks: +1, -0

If the two lines

\frac{x-1}{2} = \frac{1-y}{-a} = \frac{z}{4}

\frac{x-3}{1} = \frac{2y-3}{4} = \frac{z-2}{2}

are perpendicular, then the value of a is equal to

– 4
5
– 5
4
– 2

Solution:

Given equation of lines,

\; \frac{x-1}{2} = \; \frac{1-y}{-a} = \; \frac{z}{4}

\Rightarrow \; \; \; \frac{x-1}{2} = \; \frac{y-1}{a} = \; \frac{z-0}{4}

.....(i)

and

\; \; \; \frac{x-3}{1} = \; \frac{2y-3}{4} = \; \frac{z-2}{2}

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

.....(ii)

Here,

a_1 = 2, b_1 = a, c_1 = 4

a_2 = 1, b_2 = 2, c_2 = 2

since, lines (i) and (ii) are perpendicular,

\therefore \; \; a_1 a_2 + b_1 b_2 + c_f c_2 = 0

\Rightarrow \; \; 2 \times 1 + a \times 2 + 4 \times 2 = 0

\Rightarrow \; \; 2 + 2a + 8 = 0

\Rightarrow \; \; 2a + 10 = 0

\Rightarrow \; \; 2a = -10

\Rightarrow \; \; a = -5

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