# Prove that the quadrilateral abcd with vertices at the points A (1,3,2), B (0,2,4), C (1,1,4), D (2,2,2)

**Prove that the quadrilateral abcd with vertices at the points A (1,3,2), B (0,2,4), C (1,1,4), D (2,2,2) is a parallelogram. Calculate cosA.**

Given a quadrilateral ABCD with vertices at the points A (1,3,2), B (0,2,4), C (1,1,4), D (2,2,2), we need to prove that this is a parallelogram and calculate cos A.

To prove that a given quadrilateral is a parallelogram, we need to know that a quadrilateral whose diagonals intersect and the intersection point is halved is a parallelogram. With the help of this we prove. To do this, you need to find the coordinates of the middle from cutting AC and BD.

Coordinates of the middle of the segment AC:

x = (1 + 1) / 2 = 1; y = (3 + 1) / 2 = 2; z = (2 + 4) / 2 = 3.

BD segment midpoint coordinates:

x = (0 + 2) / 2 = 1; y = (2 + 2) / 2 = 2; z = (4 + 2) / 2 = 3.

It can be seen that the coordinates of the midpoints of the segments are the same. It follows from this that the quadrilateral ABCD is a parallelogram.

Now, we need to find cos A. It is necessary to compose AB and AC.

AB (- 1; – 1; 2) and AC (0; – 2; 2).

Let’s find the lengths and the dot product of the vectors:

AB • AC = (- 1) • 0 + (- 1) • (- 2) + 2 • 2 = 6.

Find the modulus of the vector AB.

AB = (1 + 1 + 4) ^ (1/2) = 6 ^ (1/2).

Find the module of the vector AD.

AD = (0 + 4 + 4) ^ (1/2) = 8 ^ (1/2)

Find the angle between the vectors:

cos A = 6 / (6 ^ (1/2) * 8 ^ (1/2)) = (3 ^ (1/2)) / 2.

Answer: cos A = (3 ^ (1/2)) / 2 and A = 30 0.