# FP3: equation of a planeWatch

#1
I am doing part b,

So, I was watching exam solutions to do this and he always started with a point on the plane.

So, the 2 lines intersect, and that's how I got a = 1 for part a.

I am under the impression that the position vector in the equations of the lines, L1 are L2 (presented by the blue and green circles respectively), are also points on the plane - therefore the vector: (1i + -1j +2z) - (1i + 4j + 0z) = parallel to the plane. the dot product of this vector (the red vector in the diagram) and the vector n = 0?

where am I going wrong?
Last edited by Maths&physics; 1 week ago
0
#2
or are the vectors on the plane?
0
#3
someone please tell me where I am going wrong....
0
1 week ago
#4
Firstly, you've interpreted the question wrong, evidenced by your sketch.

The plane contains the lines l1 and l2, so this means that not only are the lines parallel to the plane, they lie in the plane.

This makes your subtraction step unnecessary, because any point on the plane can be used as a position vector.

An equation is r.n=a.n which you can easily change into the required form.

Remember a (point on the plane) is easy because any point on the lines given is a point on the plane.
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