# Vectors and planes

• January 21st 2008, 04:10 AM
geton
Vectors and planes
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.

• January 21st 2008, 06:04 AM
earboth
Quote:

Originally Posted by geton
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.

Hello,

the equation of the first line is:
$
\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: $
\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:

$\cos(\theta) = \frac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}$
$
\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|}$
= $\frac{-15}{3\sqrt{3} \cdot 5\sqrt{3}} = -\frac{1}{3}$ ........ that means: $\theta \approx 109.47^\circ$

I don't know which method you use to calculate the point of intersection. I've got $r = -\frac13$ and $s = -\frac25$ produces the intersection point (1, 1, 2)