First compute the magnitudes of vectors A and B. For A, it is:
|A| = sqrt(144 + 49 + 25) = 14.765
For B,
|B| = sqrt(4 + 16 + 9) = 5.385
Then compute the dot product of A and B.
A*B = 28 - 24 -15 = -11
The value of cosine of the angle between the vectors is
A*B/|A||B| = -11/79.5 = -0.1383
The angle is 97.9 degrees
The cross product of A and B is most easily calculated with a determinant.
AxB =
|i j k|
|Ax Ay Az|
|Bx By Bz|
If you don't understand the notation, look up vector cross products with Google.
Given VectorA=12i+7j-5k and Vector B= 2i-4j+3k
1)Find angle theta between vector A and vector B?
2)Fine VectorB x VectorA
How would I go about doing these problems ?
1 answer