# Math Help - Parallelogram with Vectors

1. ## Parallelogram with Vectors

Hello! Im writing a linear Algebra test in January and im currently practicing. But my problem is, that its hard to verify my

results so id thought i could post here. Please tell me if its not appropiate. Besides that, im still not ablle to solve all

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

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]
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

2. 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
to #2): You don't need the equation of the plane to find the coordinates of point C.
You know that the mean of the position vectors of A and C is the position vector of M. Let $C(x_C, y_C, z_C)$ then

$\frac12([3, -2, 3] + [x_C, y_C, z_C]) = [6,3,7]$

You'll get C(9, 8, 11).

to #3): The orientation of the plane in 3-D is produced by the direction vectors of the plane. Therefore

$\vec n = [1,2,-3] \times[-2,-3,-6]$

Where from do you know when a vector is pointing down?

to #4):

I've got $\overrightarrow{AB} = [4,0,1]-[3,-2,3] = [1,2,-2]$
and
$\overrightarrow{AC}=[9,8,11]-[3,-2,3]=[6,12,8]$