Skewed 3D Lines (C4 Vectors) Watch

Freerider101
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If there were two 3D lines in form of vector equations which are skew. How would you workout the distance between the lines where one passes over the other?
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dknt
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(Original post by Freerider101)
If there were two 3D lines in form of vector equations which are skew. How would you workout the distance between the lines where one passes over the other?
The distance between any point on each line? Or the shortest distance between the 2 lines (i.e. the perpendicular distance).

EDIT Ignore me, just reread your question. Are you given the two points of where they pass over? If not there is a formula.. but I learnt it in FP3...
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Freerider101
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(Original post by dknt)
The distance between any point on each line? Or the shortest distance between the 2 lines (i.e. the perpendicular distance).

EDIT Ignore me, just reread your question. Are you given the two points of where they pass over? If not there is a formula.. but I learnt it in FP3...
Okay thats. Well the coordinates given are: (0, 3.3, 2.4) and (3,1.3,1.9). So I get the vector equation

r = \begin{pmatrix} 0 \\ 3.3 \\ 2.4 \end{pmatrix}+ t \begin{pmatrix} 3 \\-2\\-0.5 \end{pmatrix}

and (0.7,0,2.3) and (1.5, 4, 1.5). And I get  r = \begin{pmatrix} 0.7 \\ 0 \\-0.8 \end{pmatrix} + t \begin{pmatrix} 0.8 \\ 4\\ -0.8 \end{pmatrix}

Now I'm hoping that it is possible to work out the coordinates by inspection since I haven't got a formula to do so (I'm on OCR).
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dknt
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(Original post by Freerider101)
Okay thats. Well the coordinates given are: (0, 3.3, 2.4) and (3,1.3,1.9). So I get the vector equation

r = \begin{pmatrix} 0 \\ 3.3 \\ 2.4 \end{pmatrix}+ t \begin{pmatrix} 3 \\-2\\-0.5 \end{pmatrix}

and (0.7,0,2.3) and (1.5, 4, 1.5). And I get  r = \begin{pmatrix} 0.7 \\ 0 \\-0.8 \end{pmatrix} + t \begin{pmatrix} 0.8 \\ 4\\ -0.8 \end{pmatrix}

Now I'm hoping that it is possible to work out the coordinates by inspection since I haven't got a formula to do so (I'm on OCR).
Hmm, I'm on OCR as well, and finding the shortest distance between 2 skew lines is on Vectors for FP3. Do you mind posting the question?
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Freerider101
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(Original post by dknt)
Hmm, I'm on OCR as well, and finding the shortest distance between 2 skew lines is on Vectors for FP3. Do you mind posting the question?
A student displays here birthday cards on strings which she has pinned to opposite walls of her room, whose floor measures 3m by 4m. Relative to one corner of the room, the coordinates of the ends of the first string are (o,3.3,2.4) and (3,1.3,1.9) in metre units. The coordinates of the ends of the second string are (0.7, 0, 2.3) and (1.5, 4, 1.5). Find the difference in the heights of the two strings where one passes over the other.

Thanks in advance.
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dknt
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(Original post by Freerider101)
A student displays here birthday cards on strings which she has pinned to opposite walls of her room, whose floor measures 3m by 4m. Relative to one corner of the room, the coordinates of the ends of the first string are (o,3.3,2.4) and (3,1.3,1.9) in metre units. The coordinates of the ends of the second string are (0.7, 0, 2.3) and (1.5, 4, 1.5). Find the difference in the heights of the two strings where one passes over the other.

Thanks in advance.
I think I possibly may have a way...

At the point where they cross over each other, the x and y co-ordinates will be the same, only their z co-ordinates will differ (the height at that point).

We have the vector equations as,

\mathbf{r}_1 = \begin{pmatrix} 0 +3s \\ 3.3 -2s \\ 2.4-0.5s \end{pmatrix}

and

\mathbf{r}_2 = \begin{pmatrix} 0.7 +0.8t \\ 0 +4t \\ 2.3 -0.8t \end{pmatrix}

We know the co-efficients of \mathbf{i} and \mathbf{j} will be the same, so we can solve them simultaneously. You will then obtain a value of s and a value for t, which you can find 2 sets of x, y and z values, for each line. Then it's just a case of finding the distance between those two points.
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Freerider101
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(Original post by dknt)
I think I possibly may have a way...

At the point where they cross over each other, the x and y co-ordinates will be the same, only their z co-ordinates will differ (the height at that point).

We have the vector equations as,

\mathbf{r}_1 = \begin{pmatrix} 0 +3s \\ 3.3 -2s \\ 2.4-0.5s \end{pmatrix}

and

\mathbf{r}_2 = \begin{pmatrix} 0.7 +0.8t \\ 0 +4t \\ 2.3 -0.8t \end{pmatrix}

We know the co-efficients of \mathbf{i} and \mathbf{j} will be the same, so we can solve them simultaneously. You will then obtain a value of s and a value for t, which you can find 2 sets of x, y and z values, for each line. Then it's just a case of finding the distance between those two points.
Okay thanks a lot. I was just wondering how you know its the z coordinate that differs? Was it because the z axis is taken to be vertical and since the questions about a room it was easy enough to deduce?
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