Prove that u=5i-4j and v=2i+3j are closing an obtuse angle.

If the angle between the vectors u and v is obtuse, then
the value of cosine of the angle between u and v must be within the interval
(-1,0).

We'll calculate the cosine of the angle from the
dot product between `vecu ` and `vecv` .

We'll recall the
formula that gives the dot product of `vecu`  and `vecv`
:

`vecu` *`vecv` = |`vecu|` *|`vecv` |*cos(`vecu,vecv`
)

We'll calculate the product of
vectors:

`vecu*vecv =
(5i-4j)(2i+3j)`

`vecu*vecv = 5*2*veci^2 + 5*3*veci*vecj -
4*2*vecj*veci - 4*3*vecj^2`

`veci^2 = veci*veci =
|veci|*|veci|*cos 0`

But `|veci| = |vecj| =
1`

`veci*vecj = |veci|*|vecj|*cos90 =
0`

`vecu*vecv = 10 - 12 =
-2`

`|vecu| = sqrt(5^2 +
4^2)`

`|vecu| =
sqrt41`

`|vecv| = sqrt(2^2 +
3^2)`

`|vecv| = sqrt13`

We'll
calculate the cosine between the vectors u and v:

cos(`vecu
, vecv` ) = `(vecu*vecv)/(|vecu||vecv|)`

`cos(vecu,vecv) =
-2/sqrt533 < 0`

Since the value of
cosine angle is between the interval (-1,0) the angle closed by the vectors u and v is
obtuse.