Originally Posted by

**coobe** ...

I have a Parallelogram given by the Points A,B,C,D created in counter-clockwise direction. The middle intersection of the

Lines AC and BD is the point M.

I have the following Points given:

A(3,-2,3)

B(4,0,1)

M(6,3,7)

Question 1: What is the length of AB ?

i have: AB = [4-3;0+2;1-3] = [1;2;-2]

|AB| = square root of 1+4+4 = 3 which is my lenght **<<<<< OK**

QUestion 2: Parametric euqation of the Plane ABM and which parameters result in the Position vector of C ?

my attempt: E(c,k) = [3;-2;3] + c[4-3;0+2;1-3] + k[4-6;0-3;1-7] = [3;-2;3] + [c;2c;-2c]+[-2k;-3k;-6k] **= [3,-2,3] + c[1,2,-2] + k[-2,-3,-6]**

which would describe my Plane ABM in parametric form

Now position vector of C with the parameters of c,k in the Plane equation.

My thinking: the length of AC is twice as long as the length AM. So i came to the conclusion my parameters should be

c = 0 since i dont need to go in the direction of point B

and d = 2 since its twice as long. Really not sure how to solve that more delicate

Question 3: Find the vector v perpendicular to the Plane ABM, such that it is pointing downwards

I think i need something like a Normalvector found by a crossproduct. But im confused since i thought i needed only 2 vectors

for that....

Question 4: Find the cosine of the angle between the lines AB and AC at the point A

cos(phi) = [ (v1 * v2) / (|v1| x |v2|) ]

v1 = AB = [3;-2;3] +c [4;0;1]

v2 = AC = A2M

not sure about that last one