Find the point of intersection of the line through the points (2, 0, 1) and (-1, 3, 4) and the line through the points (-1, 3, 0) and (4, -2, 5). Calculate the acute angle between the two lines.
How can I solve this? Please help me.
Hello,
the equation of the first line is:
$\displaystyle
\vec x = \left(\begin{array}{c}2\\0\\1\end{array}\right) + r \cdot \left(\left(\begin{array}{c}2\\0\\1\end{array}\rig ht) - \left(\begin{array}{c}-1\\3\\4\end{array}\right)\right)$ ........and the equation of the 2nd line is: $\displaystyle
\vec x = \left(\begin{array}{c}-1\\3\\0\end{array}\right) + s \cdot \left(\left(\begin{array}{c}-1\\3\\0\end{array}\right) - \left(\begin{array}{c}4\\-2\\5\end{array}\right)\right)$
Take the direction vectors to calculate the angle according to the formula:
$\displaystyle \cos(\theta) = \frac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}$
$\displaystyle
\cos(\theta)=\frac{\left(\begin{array}{c}3\\-3\\-3\end{array}\right) \cdot \left(\begin{array}{c}-5\\5\\-5\end{array}\right)} {\left|\left(\begin{array}{c}3\\-3\\-3\end{array}\right) \right| \cdot \left| \left(\begin{array}{c}-5\\5\\-5\end{array}\right)\right|}$ = $\displaystyle \frac{-15}{3\sqrt{3} \cdot 5\sqrt{3}} = -\frac{1}{3}$ ........ that means: $\displaystyle \theta \approx 109.47^\circ$
I don't know which method you use to calculate the point of intersection. I've got $\displaystyle r = -\frac13$ and $\displaystyle s = -\frac25$ produces the intersection point (1, 1, 2)