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Mathematics Question on Three Dimensional Geometry

The value of λ\lambda for which the lines 1−x3=y−22λ=z−32=x−13λ=y−11=6−z7\frac{1-x}{3} = \frac{y-2}{2\lambda}=\frac {z-3}{2} = \frac{x-1}{3\lambda} =\frac{y-1}{1} =\frac {6-z}{7} are perpendicular to each other is

A

-1

B

-2

C

1

D

2

Answer

-2

Explanation

Solution

Given, 1−x3\frac{1-x}{3}
=y−22λ=\frac{y-2}{2\lambda}
=z−32…(i)=\frac{z-3}{2} \ldots\left(i\right)
and x−13λ=y−11=6−z7…(ii)\frac{x-1}{3\lambda} =\frac{y-1}{1} =\frac{6-z}{7} \ldots\left(ii\right)
Eqs.(i) \left(i\right) and (ii)\left(ii\right), can be written as
x−1−3=y−22λ=z−32\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}
and x−13λ=y−11 \frac{x-1}{3\lambda} =\frac{y-1}{1}
=z−6−7=\frac{z-6}{-7}
Since, both are perpendicular to each other, then
(−3)(3λ)+(2λ)(1)+(2)(−7)=0\left(-3\right)\left(3\lambda\right)+\left(2\lambda\right)\left(1\right)+\left(2\right)\left(-7\right)=0
⇒−9λ+2λ−14=0\Rightarrow -9\lambda+2\lambda-14=0
⇒7λ=−14\Rightarrow 7\lambda=-14
⇒λ=−2\Rightarrow \lambda=-2