If the lines $\frac{x-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-1}{4}$ and $\frac{x-3}{1}=\frac{\mathrm{y}-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ interect, then the value of the $\mathrm{k}$ is

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Aicharya · Accepted Answer

Let $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}ight) \equiv(1,-1,1)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}ight) \equiv(3, \mathrm{k}, 0)$ Let $\left(\mathrm{a}_{1}, \mathrm{~b}_{1}, \mathrm{c}_{1}ight) \equiv(2,3,4)$ and let $\left(\mathrm{a}_{2}, \mathrm{~b}_{2}, \mathrm{c}_{2}ight) \equiv(1,2,1)$ Since, the lines intersect $\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \ a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2}\end{array}ight|=0$ $\left|\begin{array}{ccc}2 & k+1 & -1 \ 2 & 3 & 4 \ 1 & 2 & 1\end{array}ight|=0$ $2(3-8)-(k+1)(2-4)-1(4-3)=0$ $-10+2 k+2-1=0 \Rightarrow k=\frac{9}{2}$

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

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