If the shortest distance between the straight lines 3(x-1)=6(y-2)=2(z-1) and 4(x-2)=2(y-)=…

Question

If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda)=(z-3)$, $\lambda\in\mathbb R$, is $\dfrac{1}{\sqrt{38}}$, then the integral value of $\lambda$ is equal to:

Accepted Answer

$${L_1}:{{(x - 1)} \over 2} = {{(y - 2)} \over 1} = {{(z - 1)} \over 3}\overrightarrow {{r_1}}  = 2\widehat i + \widehat j + 3\widehat k$$

$${L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}}  = \widehat i + 2\widehat j + 4\widehat k$$

Shortest distance = Projection of $${\overrightarrow a }$$ on $${\overrightarrow {{r_1}}  \times \overrightarrow {{r_2}} }$$

$$ = {{\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}}  \times \overrightarrow {{r_2}} } \right)} \right|} \over {\left| {\overrightarrow {{r_1}}  \times \overrightarrow {{r_2}} } \right|}}$$

$$\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}}  \times \overrightarrow {{r_2}} } \right)} \right| = \left| {\begin{pmatrix} 1 & {\lambda  - 2} & 2  \
   2 & 1 & 3  \
   1 & 2 & 4  \

\end{pmatrix} } \right| = \left| {14 - 5\lambda } \right|$$

$$\left| {\overrightarrow {{r_1}}  \times \overrightarrow {{r_2}} } \right| = \sqrt {38} $$

$$\therefore$$ $${1 \over {\sqrt {38} }} = {{\left| {14 - 5\lambda } \right|} \over {\sqrt {38} }}$$

$$ \Rightarrow \left| {14 - 5\lambda } \right| = 1$$

$$ \Rightarrow 14 - 5\lambda  = 1$$ or $$14 - 5\lambda  =  - 1$$

$$ \Rightarrow \lambda  = {{13} \over 5}$$ or 3

$$\therefore$$ Integral value of $$\lambda$$ = 3.

Solutions · 1