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Two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point, if $k$ is equal to
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Verified Answer
The correct answer is:
$\frac{9}{2}$
$\begin{array}{l}
\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=r(\text { say }) \\
\Rightarrow x=2 r+1, y=3 r-1, z=4 r+1
\end{array}$
Since, the two lines intersect.
So, putting above values in second line, we get
$\begin{array}{l}
\frac{2 r+1-3}{1}=\frac{3 r-1-k}{2}=\frac{4 r+1}{1} \\
2 r-2=4 r+1
\end{array}$
$\begin{aligned} \Rightarrow & \mathrm{r}=-3 / 2 \\ & \text { Also } 3 \mathrm{r}-1-\mathrm{k}=8 \mathrm{r}+2 \\ \Rightarrow & \mathrm{k}=-5 \mathrm{r}-3=\frac{15}{2}-3=\frac{9}{2} \end{aligned}$
\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=r(\text { say }) \\
\Rightarrow x=2 r+1, y=3 r-1, z=4 r+1
\end{array}$
Since, the two lines intersect.
So, putting above values in second line, we get
$\begin{array}{l}
\frac{2 r+1-3}{1}=\frac{3 r-1-k}{2}=\frac{4 r+1}{1} \\
2 r-2=4 r+1
\end{array}$
$\begin{aligned} \Rightarrow & \mathrm{r}=-3 / 2 \\ & \text { Also } 3 \mathrm{r}-1-\mathrm{k}=8 \mathrm{r}+2 \\ \Rightarrow & \mathrm{k}=-5 \mathrm{r}-3=\frac{15}{2}-3=\frac{9}{2} \end{aligned}$
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