If the line x-12=y+13=z-14 and x-31=y-k2= z1 intersect, then k is equal to:

Question

If the line $\dfrac{x-1}2=\dfrac{y+1}3=\dfrac{z-1}4$ and $\dfrac{x-3}1=\dfrac{y-k}2=\dfrac z1$ intersect, then $k$ is equal to:

Accepted Answer

Given lines in vector form are

$$\overrightarrow r  = \left( {\widehat i - \overrightarrow j  + \overrightarrow k } \right) + \lambda \left( {2\overrightarrow i  + 3\overrightarrow j  + 4\overrightarrow j } \right)$$

and $$\overrightarrow r  = \left( {3\widehat i + k\widehat j} \right) + \mu \left( {\widehat i + 2\widehat j + \widehat k} \right)$$

These will intersect if shortest distance between them $=0$

i.e.$$\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).\overrightarrow {{b_1}}  \times {\overrightarrow b _2} = 0$$

$$ \Rightarrow \left| {\begin{pmatrix} {3 - 1} & {k + 1} & { - 1}  \
   2 & 3 & 4  \
   1 & 2 & 1  \

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

$$ \Rightarrow 2\left( { - 5} \right) - \left( {k + 1} \right)\left( { - 2} \right) - 1\left( 1 \right) = 0$$

$$ \Rightarrow k = 9/2$$

Solutions · 1