Find the scalar and vector products of the vectors A and B, where
A = 2i + j + k and
B = 4i + 2j - 3k, also find the angle between A and B
-------------
Scalar or dot product = 2*4 + 1*2 + 1*-3 = 7
------
The cosine of the angle is the dot product over the magnitude product.
cos(t) = (A dot B)/(|A|*|B|)
|A| = sqrt(2^2 + 1^1 + 1^2) = sqrt(6)
|B| = sqrt(4^2 + 2^2 + 3^2) = sqrt(29)
cos = 7/sqrt(6*29)
Angle =~ 57.95º
==========================
Vector or Cross Product:
|i j k|
|2 1 1| = i*(-3-2) - j*(-6-4) + k*(4-4) = -5i + 10j
|4 2 -3|
= -5i + 10j