Question

Two lines $\frac{[x-1]}{2}=\frac{[y+1]}{3}=\frac{[z-1]}{4}$ and $\frac{[x-3]}{1}=\frac{[y-k]}{2}=\frac{z}{1}$ intersect at a point, if$k=$

• A. $\frac{2}{9}$
• B. $\frac{1}{2}$
• C. $\frac{9}{2}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C

$\frac{9}{2}$

Explanation for correct option:

Find the value of $k$

The two equations are

$\frac{[x-1]}{2}=\frac{[y+1]}{3}=\frac{[z-1]}{4}.....\left(1\right)$

$\frac{[x-3]}{1}=\frac{[y-k]}{2}=\frac{z}{1}......\left(2\right)$

Let the equation $\frac{[x-1]}{2}=\frac{[y+1]}{3}=\frac{[z-1]}{4}=r$

$\Rightarrow x=2r+1,y=3r-1,z=4r+1$

Putting the above value in the $\left(2\right)$ equation

$\frac{2r+1-3}{1}=\frac{3r-1-k}{2}=\frac{4r+1}{1}$

Taking the first and the last together

$$\begin{gathered} \frac{2r+1-3}{1}=\frac{4r+1}{1} \\ \Rightarrow 2r=-3 \\ \Rightarrow r=\frac{-3}{2} \end{gathered}$$

Taking the second and the last together

$$\begin{gathered} =\frac{3r-1-k}{2}=\frac{4r+1}{1} \\ \Rightarrow k=-5\times \left(\frac{-3}{2}\right)-3 \\ \Rightarrow k=\frac{9}{2} \end{gathered}$$

On solving the above equation

$\Rightarrow r=-\frac{3}{2}$

$\Rightarrow k=\frac{9}{2}$

Hence, the correct option is $C$.