Question
Find the direction cosines of the line through the origin which intersects each of the lines
$\displaystyle \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\displaystyle \frac{x+2}{4} = \frac{y-3}{3} = \frac{z-4}{2}$.
My Attempt
Projection of a vector connecting the two lines on the perpendicular from the required line to the given two lines is zero. Hence
\begin{equation}
\begin{vmatrix}
1&2&3\\
2&3&4\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}
\begin{equation}
\begin{vmatrix}
-2&3&4\\
4&3&2\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}
This gives me two equations
\begin{equation}
a-2b+c=0
\end{equation}
\begin{equation}
3a-10b+9c=0
\end{equation}
I get two different ratios for the direction cosines (a,b,c) as
\begin{equation}
(5,3,1),(2,3,1)
\end{equation}
Both these direction cosines give me a line intersecting only one given line.
Is my approach wrong or does this mean that there is no line that intersects both given lines and also goes through the origin?
Highly appreciate your help on this.Thanks in advance.
Best Answer
Your process and equations are correct, but I can't figure out how you arrived at your answers from them. Here's what I did: $$a-2b+c=0 {\tag 1}$$ $$3a-10b+9c=0 {\tag 2}$$ Let $\frac ac=k$, $\frac bc=l$, then, from $(1)$: $$k-2l+1=0$$ and from $(2)$: $$3k-10l+9=0$$ Solving these gives $k=2$ and $l=\frac 32$. Hence, $a=2c$ and $b=\frac {3c}{2}$. Thus, direction ratio are: $$\left(2c,\frac {3c}{2}, c \right)\equiv \left(2, \frac 32, 1 \right) \equiv (4,3,2)$$ Thus we get only one direction ratio, $(4,3,2)$. Hence equation of the required line is: $$\frac x4=\frac y3=\frac z2$$
Addendum: Since our original condition is satisfied for both intersecting and parallel lines, it is necessary to check whether this line intersects or is parallel to either of our given lines. As explained in the comments, it turns out that this line is parallel to one of the lines, and hence no line exists which intersects both.