You are wondering if might be the answer. Well, would give , , but doesn’t lie on the plane . Therefore can’t be correct.
Edit : Tutor confirmed the messy answer is correct.
In the assignment, Question 6 relies on correct answers for 4 and 5. I think I have made a small sign error somewhere, since the numbers are not typical of an assignment question otherwise.
Take a look at Q4 and Q5
Q6 requires me to find the intersection between both of those (the line L2 in Q4 and the plane in Q5)
I get down to a final bit that say 101t + 46 = 55 (to find t and then sub back in to the parametric equation)
This gives me t = 9/101
However, the fact that if the 46 was negative, it would be 101t = 55+46 would give me t = 1 is too obvious to ignore. So I think I should have 101t-46 = 55 but I can't see where I have gone wrong.
Attached is my working.
There is a small mistake in this one, the vector equation for s should be (-3,-1,2) but I wrote (-3,1,3) for some reason. However, in the system below that, you can see it is correct (ie, y = 2 -s -5t)
Lukasaurus
I have verified all these messy calculations and I found no mistake.
However the method you have used to find the equation of the plane is lengthy and ubnormal
The problems you stated are typical problems of 3D geometry for IG grade -12 students.
it would be better to get the equation of the plane finding first the normal to the plane by getting the vector product AB X AC .the rest are easy.
I have verified also the value of t=9/101 and there is nothing wrong with it.
is you substitute this value into the parametric equations and then to the equation of the plane you will find 55
therefore there is nothing wrong with that.
I wish you Good luck to the next C.I.E exams.
MINOAS