FP3: intersection of planes Watch

Maths&physics
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#21
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#21
(Original post by old_engineer)
The most direct way of finding n, the direction vector perpendicular to the plane, is to use the direction vectors of L1 and L2. n must be perpendicular to both the line direction vectors.
(Original post by ghostwalker)
In red is a vector parallel to the plane.

The rest of the line doesn't make sense in relation to that.



What do you mean by "the vector to the plane".

The dot product of n with any vector in the plane or parallel to a vector in the plane will be zero.



First one defines a point on l1 and second one is a point on l2.


Please think about what you're trying to say - I am really struggling to make sense of your posts relating to this question.
(Original post by DFranklin)
Yes (noting I haven't actually checked those are the correct direction vectors). And you need to find a solution *other* than x=y=z=0.

Do you know what the vector cross-product is?
thank all 3 of you ever so much! I got 6x + 2y 3z - 14 = 0, as my final answer and it's correct!!!!!!!
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DFranklin
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#22
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(Original post by Maths&physics)
yes, I know and thats where I got the dot product from.
If you know what a vector cross product is, the standard method is to use it to get a vector perpendicular to both direction vectors.
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Maths&physics
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#23
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(Original post by DFranklin)
If you know what a vector cross product is, the standard method is to use it to get a vector perpendicular to both direction vectors.
yes, that's what I did but I was confused whether the known points of the lines where on the plane - which they are, but not in the way I thought because I drew the lines crossing through the planes, rather than on the plane.
Last edited by Maths&physics; 1 week ago
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begbie68
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so how did you get on with part c of this question?
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Maths&physics
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#25
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(Original post by begbie68)
so how did you get on with part c of this question?
I got the right answer, thanks for asking.

.....are you offering help or would you like some?
Last edited by Maths&physics; 1 week ago
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