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