If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ have a point in common, then $k=$

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Aicharya · Accepted Answer

Given line is $ \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda $ So, $(x, y, z)$ is $(2 \lambda+1,3 \lambda-1,4 \lambda+1)$ and this point is lies on given line. This point also lies on line. $ \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} $ So, this point satisfies equation $ \begin{array}{rlrl} & & \frac{2 \lambda+1-3}{1} & =\frac{3 \lambda-1-k}{2}=\frac{4 \lambda+1}{1} \ \Rightarrow & & \frac{2 \lambda-2}{1} & =\frac{4 \lambda+1}{1} \ \Rightarrow & 2 \lambda-2 & =4 \lambda+1 \ \Rightarrow & & 2 \lambda & =-3 \ \Rightarrow & & \lambda & =-\frac{3}{2} \end{array} $ Now, $\frac{3 \lambda-1-k}{2}=\frac{2 \lambda-2}{1}$ $ \Rightarrow \frac{3 	imes-\frac{3}{2}-1-k}{2}=\frac{2 	imes-\frac{3}{2}-2}{1} $ $\begin{array}{ll}\Rightarrow & \frac{-\frac{9}{2}-1-k}{2}=\frac{-3-2}{1} \Rightarrow-\frac{11}{2}-k=-10 \ \Rightarrow & -\frac{11}{2}+10=k \Rightarrow k=\frac{9}{2}\end{array}$

If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$…

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