Find $k$ such that the angle between the vectors $(2,k)$ and $(3,5)$ is $60$ degrees

I have $2$ vectors : $U =(2,k)$ and $V = (3,5)$. I want to find the $k$ value when the angle between $U$ and $V$ is $60$ degrees.

This what I tried to do but I don't get the right answer :

$2\cdot3 + 5k = \sqrt{4+k^2} \cdot \sqrt{34} \cdot \cos60 \rightarrow 24k^2 +30.84k+5$

• The idea is correct, but with so many steps missing I'm unable to tell what went wrong. Commented Nov 11, 2013 at 7:16
• Because there so many steps , do you think it's possible to find the K value using another way ? Commented Nov 11, 2013 at 7:19

The angle between V and the X axis is $arctan(5/3) = 59 \deg.$

Therefore, if $U$ is about 1 deg. below the X axis, the angle between U and V will be 60 degrees.

Now solve:

$k/2 = tan(-1)$

and get the solution.

• k/3 = tan(-1) is -0.05 . This solution doesn't work when I try to validate . Commented Nov 11, 2013 at 8:16
• Yes, $U=(2,-0.05)$ should be a possible solution. How did you try to validate? Commented Nov 11, 2013 at 8:23
• just use the vectors on those function : |U||V|cos60 = Ux Vx +UyVy Commented Nov 11, 2013 at 8:28
• Sorry, this should of course be $k/2$. So the solution is $U=(2,-0.035)$. Now $|U||V| \cos 60 = 2 * 5.83 * 0.5 = 5.83$, and $Ux *Vx +Uy*Vy = 2*3 - 5*0.035=5.83$ Commented Nov 11, 2013 at 8:38