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FP3: intersection of planes

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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!!!!!!! :smile:

Reply 21

DFranklin

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.

Reply 22

Maths&physics

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.