# The vectors`veca`and`vecb`have magnitude of 6 and 8.The angle between a and b is obtuse and magnitude of a x b is 16. Find the angle between a and b.

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### 2 Answers

You need to remember that evaluating the magnitude of cross product of vectors `bar a` and `bar b` yields:

`|bar a X bar b| = |bar a|*|bar b|*sin alpha`

The problem provides the magnitude of cross vector and the magnitudes of each vector `bar a` and `bar b` , hence you should substitute 16 for `|bar a X bar b|` and 6 for `|bar a|` and 8 for `|bar b|` such that:

`16 = 6*8 sin alpha`

Dividing by 16 both sides yields:

`1 = 3 sin alpha =gt sin alpha = 1/3 =gt alpha = arcsin (1/3)`

`alpha~~ 19.26^o`

The problem provides the information that angle between vectors `bar a` and `bar b` is obtuse, hence `alpha~~ 180^o - 19.26^o ~~ 160.74^o` .

**Hence, evaluating the obtuse angle between vectors bar a and bar b, under given conditions, yields `alpha~~ 160.74^o` .**

The cross product of two vectors `vec a` and` vec b` is `|vec a|*|vec b|*sin theta` .

`vec a` has a magnitude of 6 and `vec b` has a magnitude of 8. The cross product of the two is 16.

`|vec a|*|vec b|*sin theta` = 16

=> 6*8*`sin theta` = 16

=> `sin theta` = (1/3)

`theta = sin^-1(1/3)`

As the angle between the vectors is obtuse, `theta` = 160.52 degrees.

**The angle between `vec a` and `vec b` is 160.52 degrees.**