Question

The complete set of values of $\alpha $ for which the lines $\frac{x - 1}{2}=\frac{y - 2}{3}=\frac{z - 3}{4}$ and $\frac{x - 3}{2}=\frac{y - 5}{\alpha }=\frac{z - 7}{\alpha + 2}$ are coplanar is

• A. $\left\{2 , 3\right\}$
• B. $\left\{0 , 3\right\}$
• C. $\left[- 2 , 3\right]$
• D. $R$

Answer 1

Answer: (D)

Any point on the first line is $\left(1 + 2 \lambda , 2 + 3 \lambda , 3 + 4 \lambda ,\right)$
Any point on the second line is $\left(3 + 2 \mu , 5 + \alpha \mu , 7 + 2 \mu + \alpha \mu \right)$
Now, as they are intersecting $1+2\lambda =3+2\mu $ and $2+3\lambda =5+\alpha \mu $
So, $\lambda =\mu +1,$ hence $2+3\left(\mu + 1\right)=5+\alpha \mu $
$\Rightarrow \left(\alpha - 3\right)\mu =0$
Hence for $\lambda =1,\mu =0,$ the point $\left(3 , 5 , 7\right)$ lies on both the lines irrespective of $\alpha .$
Hence, $\forall \alpha \in R$ lines are concurrent.