1
$\begingroup$

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 :

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

$\endgroup$
2
  • 2
    $\begingroup$ The idea is correct, but with so many steps missing I'm unable to tell what went wrong. $\endgroup$
    – roman
    Nov 11, 2013 at 7:16
  • $\begingroup$ Because there so many steps , do you think it's possible to find the K value using another way ? $\endgroup$
    – Malikdz
    Nov 11, 2013 at 7:19

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ k/3 = tan(-1) is -0.05 . This solution doesn't work when I try to validate . $\endgroup$
    – Malikdz
    Nov 11, 2013 at 8:16
  • $\begingroup$ Yes, $U=(2,-0.05)$ should be a possible solution. How did you try to validate? $\endgroup$ Nov 11, 2013 at 8:23
  • $\begingroup$ just use the vectors on those function : |U||V|cos60 = Ux Vx +UyVy $\endgroup$
    – Malikdz
    Nov 11, 2013 at 8:28
  • 1
    $\begingroup$ 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 $Ux *Vx +Uy*Vy = 2*3 - 5*0.035=5.83$ $\endgroup$ Nov 11, 2013 at 8:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .