If the lines (x-1)/(2)=(2-y)/(-3)=(z-3)/( ) and (x-4)/(5)=(y-1)/(2)=(z)/( ) intersect, the…

Question

If the lines $\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha }$ and $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta }$ intersect, then the magnitude of the minimum value of $8\alpha \beta $ is \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 18

Accepted Answer

Given, lines

$\begin{gathered} \frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha } \ \Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\alpha }=\lambda .........(i) \end{gathered}$

Any point on the line (i)

$x=2\lambda +1,y=3\lambda +2,z=\alpha \lambda +3$

and line $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta }=\mu $ ............(ii)

Any point on line (ii)

$\Rightarrow x=5\mu +4,y=2\mu +1,z=\beta \mu $

Since, given lines intersects

$\begin{gathered} 	herefore 2\lambda +1=5\mu +4..........(iii) \ 3\lambda +2=2\mu +1............(iv) \ \alpha \lambda +3=\beta \mu ..........(iv) \end{gathered}$

On solving (iii) and (iv), we get

$\lambda =-1,\mu =-1$

On putting value of $\lambda $ and $\mu $ in (v), we get

$\begin{gathered} \alpha (-1)+3=-\beta \ \Rightarrow \alpha =\beta +3 \end{gathered}$

Now,

$\begin{gathered} 8\alpha \beta =8(\beta +3)(\beta ) \ =8({\beta }^{2}+3\beta ) \ =8({\beta }^{2}+3\beta +\frac{9}{4}-\frac{9}{4}) \ =8{{(\beta +\frac{3}{2})}^{2}-\frac{9}{4}} \end{gathered}$

$=8{(\beta +\frac{3}{2})}^{2}-18$

Here, minimum value $=-18$

$	herefore $ Magnitude of the minimum value of $8\alpha \beta $ is 18 .

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