OK so the question to blame is
if the triangle PQRR has vertices
P(-1,2,3), Q(1,1,5), and R(0,4,5) then what is the angle at P (I already know the right answer but my working out doesn't lead to it and I can't see where I'm going wrong)

OK

so I work out PR and PQ to find the vectors so I can use the dot product
PR=a= (1i,2j,2k) and PQ=b=(2i,-1j,2k)

so using the dot product I get a.b = 4

so I find ab=15? This is what I'm not entirely sure on

therefore (a.b/ab)=cos(theta)

so cos^-1(4/15) = theta

I get 74.5 but that is the wrong answer

the right answer I'm supposed to be getting is 63.6
but as to how is a mystery to me :confused:
any help is greatly appreciated

Aha figured it out its k, on to the next problem :D

You know, another way to go about this is to notice this is an isosceles triangle.

It has side lengths 3 and base = sqrt(10).

2sin^{-1}(\frac{\sqrt{10}}{6})=63.61 \;\ degrees

Yea I don't know how to recognise an isocelese triangle from that :o
Plus that's the way we've been shown in class so presumably that's how they want us to work them out :)

Use Pythagoras to find the distance between the points. Doing this, we see two of the distances are 3. That means it is isosceles.

Reet gotchya :)