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 thesides of the triangleare sqrt(5), sqrt(38) and sqrt(17)

Area of parallelogram= the magnitude of the cross product ofPQandPR

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

|PQcrossPR| = sqrt(21)

Thearea of the triangleis therefore1/2 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.