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Two straight lines are given:

$$ \left( \begin{array}{c} 1\\ 1\\ 4 \end{array} \right) + t \left( \begin{array}{c} 4\\ 1\\ 1 \end{array} \right) $$

and

$$ \left( \begin{array}{c} 5\\ 5\\ 2 \end{array} \right) + s \left( \begin{array}{c} 0\\ 1\\ -1 \end{array} \right) $$

I want to calculate the angle between them (which should be simple, arccos of the scalar product over the product of their norms). Since the angle should not change no matter which $s$ and $t$ we choose, I took arbitrary parameters $t = 2$ and $s = 1$ and calculated first just the scalar product of $(9,3,6)$ and $(5,6,1)$ - that is clearly non-zero, though the solutions state that the angle between the lines should be 90 degrees - what am I doing wrong?

Thanks

Jay Zha
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Pugl
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  • the inner product of (4, 1, 1) and (0, 1, -1) is zero (40 + 11 + 1*-1 = 0).So these lines are orthogonal –  Mar 26 '17 at 14:17
  • why is it wrong if I add the parameterized direction to the starting point as a vector? – Pugl Mar 26 '17 at 14:18
  • there is no need for that. The direction of the lines is given by (4,1,1) and (0, 1, -1). You just calculated points on the line –  Mar 26 '17 at 14:20

1 Answers1

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So the problem is that once you fix $s$ and $t$, you are fixing two points on the line, and the inner product you get over the two points are related to the angle of the vectors starting from origin $(0,0)$ to those two points. Thus is not correct.

You want to take the inner product of the vector that is of the same direct of the line. And in this case it should be $(4,1,1)$ and $(0,1,-1)$

Intuitively, the lines are defined by one point and a direction, and $t$, $s$ just give points the freedom to move along the direction, which will conform a line.

Jay Zha
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    Ok I think I get it..my approach hence would be correct if the straigh lines would go through the origin, right? But since that is not a must, my approach is generally wrong – Pugl Mar 26 '17 at 15:16
  • @Pegah Yes, you are right. that is the case (+1) – Jay Zha Mar 26 '17 at 15:18