Vectors and planes

**geton** · January 21st 2008, 03:10 AM

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.

**earboth** · January 21st 2008, 05:04 AM

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.

How can I solve this? Please help me.

Hello,

the equation of the first line is:
$<br /> \vec x = \left(\begin{array}{c}2\\0\\1\end{array}\right) + r \cdot \left(\left(\begin{array}{c}2\\0\\1\end{array}\rig ht) - \left(\begin{array}{c}-1\\3\\4\end{array}\right)\right)$ ........and the equation of the 2nd line is: $<br /> \vec x = \left(\begin{array}{c}-1\\3\\0\end{array}\right) + s \cdot \left(\left(\begin{array}{c}-1\\3\\0\end{array}\right) - \left(\begin{array}{c}4\\-2\\5\end{array}\right)\right)$

Take the direction vectors to calculate the angle according to the formula:

$\cos(\theta) = \frac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}$
$<br /> \cos(\theta)=\frac{\left(\begin{array}{c}3\\-3\\-3\end{array}\right) \cdot \left(\begin{array}{c}-5\\5\\-5\end{array}\right)} {\left|\left(\begin{array}{c}3\\-3\\-3\end{array}\right) \right| \cdot \left| \left(\begin{array}{c}-5\\5\\-5\end{array}\right)\right|}$ = $\frac{-15}{3\sqrt{3} \cdot 5\sqrt{3}} = -\frac{1}{3}$ ........ 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)

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