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Math Help - Vectors and planes

  1. #1
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    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.

    How can I solve this? Please help me.
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  2. #2
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    Quote Originally Posted by geton View Post
    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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