Finding Position Vectors and Co-planarity

[mariechap89]

a) Masses, m1,m2,...mk are at points P1,P2,...Pk with position vestors P1,P2,...Pk (these should be in bold) respectively. The position vector g of the centre mass is given by: (m1+m2+...mk)g=m1,p1+m2p2+...+mkpk

Find the position vector of the centre of mass of the masses 3,4, and 5kg placed at:
P1=(1,-1,1), P2=(2,-1,1) and P3=(-1,0,2) respectively.

Not sure how to do this

b) Determine the value of lambda that will make the following points coplanar:
(1,0,-3), (1,1,-2) and (lambda,-1,0).

Not sure how to do this either

c) Find a unit vector normal to the curve 2cosx-3e^x=y^2-1 at the point where x=-pi/4
Not sure how to do this as well

Any help would be great, even if its just a point in the right direction to start with

 

[CompuChip]

You have been given the equation (let me work it out in LaTeX):

(m_1 + \cdots + m_k) \vec g = m_1 \vec p_1 + \cdots + m_k \vec p_k
from which you can solve for the position vector g:
\vec g = \frac{ m_1 \vec p_1 + \cdots + m_k \vec p_k }{m_1 + \cdots + m_k}

All you have to do is write it down for your specific case where k = 3 and plug in the values.

For b, what does it mean if three points are co-planar?

For c, even if you have no clue, you can calculate the y-coordinate of the point they are talking about. Then you can look up how you can find a normal vector to a curve, at a given point (it has to do with the tangent vector to the curve).