If the straight lines x-1k=y-22=z-33 and x-23=y-3k=z-12 intersect at a point, then the int…

Question

If the straight lines $\dfrac{x-1}k=\dfrac{y-2}2=\dfrac{z-3}3$ and $\dfrac{x-2}3=\dfrac{y-3}k=\dfrac{z-1}2$ intersect at a point, then the integer $k$ is equal to

Accepted Answer

The two lines intersect if shortest distance between them is zero $i.e.$

$${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$$

$$ \Rightarrow \left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$$

where $${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k,{\overrightarrow b _1} = k\widehat i + 2\widehat j + 3\widehat k$$

$$\,\,\,\,\,\,\,{\overrightarrow a _2} = 2\widehat i + 3\widehat j + \widehat k,\,\,{\widehat b_2} = 3\widehat i + k\widehat j + 2\widehat k$$

$$ \Rightarrow \left| {\begin{pmatrix} 1 & 1 & { - 2}  \
   k & 2 & 3  \
   3 & k & 2  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0$$

$$ \Rightarrow  - 2{k^2} - 5k + 25 = 0 \Rightarrow k =  - 5$$ $$\,\,\,\,$$ or $$\,\,\,\,$$ $${5 \over 2}$$

As $k$ is an integer, therefore $k=-5$

Solutions · 1