Proceeding with the method you have described is taking a very long time and leaving me to the power of 4 that are very hard to work out! Especially as this question is form a non-calculator exam.
I think there must be and easier way to work this out.
The way I had done it, is by working out
and then establishing a relationship between the line and the position vector (2,1,2)
This is possible since before i had proved that this position vector was perpendicular to both and hence it must be parallel to the line
with this I equated the i and k components of
and doubled the component j of
However the numbers I'm getting seem really odd.
Hello maggiecI agree with your working, and also your method - which is to say that this vector is parallel to .
I have done this, and I also get rather nasty numbers. I get the following equations for and :
andwhich give (if my working is correct):
andIs that what you make it?
Grandad
Yeah!! that's exactly what i get!
But from the answers I've been giving for this question, putting this back into the equation for each line doesn't give me the correct and vectors.
These are meant to be: and
Sudharaka:
I get to a point where 202 - (56/3) + (3304/39) + (1981/39) + 234.((26 -118 - 70.75)/78)^2 - 14 = 0
that squared after the bracket makes me think I will get terms although I haven't expanded on it yet.