Vector Problem. Finding the angles and sides of a triangle in 3d as well as the area.

Problem:

Find all angles and the lengths of all the sides in the triangle with the corners P(1,2,3) Q(0,2,1) R(-2,4-2). Hence compute its area.

Solution so far:

Vectors (**u **= u2 - u1)

**PQ = **(-1,0,-2), **PR **= (-3,2,-5)

**QP **= (1,0,2), **QR =**(-2,2-3)

**RQ **= (2,-2,3), **RP **= (3,-2,5)

Lengths of vectors |**u**| = sqrt(u1^2 + u2^2 + u3^2)

hence,

|**PQ**| = sqrt(5)

|**PR**| = sqrt(38)

|**QP**| = sqrt(5)

|**QR**| = sqrt(17)

|**RQ**| = sqrt(17)

|**RP**| = sqrt(38)

Therefore the **sides of the triangle** are sqrt(5), sqrt(38) and sqrt(17)

**Area of parallelogram** = the magnitude of the cross product of **PQ **and **PR**

PQ cross PR = (4, 1 ,-2)

|**PQ** cross **PR**| = sqrt(21)

The **area of the triangle** is therefore **1/2 sqrt(21)**

Angles between vectors

PQ and PR => cosine(a) = (**PQ**.**PR**)/(|**PQ**||**PR**|)

therefore a = cos^-1 (**PQ**.**PR**)/(|**PQ**||**PR**|)

Now if I calculate the angle between all the vectors then the sum of those angles should be 180 degrees, right...

Are there any problems with my workings?

Basically, am I doing anything wrong?

Feedback would be greatly appreciated.