Solution :
Let [tex]$v_0$[/tex] be the unit vector in the direction parallel to the plane and let [tex]$F_1$[/tex] be the component of F in the direction of [tex]v_0[/tex] and [tex]F_2[/tex] be the component normal to [tex]v_0[/tex].
Since, [tex]|v_0| = 1,[/tex]
[tex]$(v_0)_x=\cos 60^\circ= \frac{1}{2}$[/tex]
[tex]$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$[/tex]
Therefore, [tex]v_0 = \left<\frac{1}{2},\frac{\sqrt 3}{2}\right>[/tex]
From figure,
[tex]|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3[/tex]
We know that the direction of [tex]F_1[/tex] is opposite of the direction of [tex]v_0[/tex], so we have
[tex]$F_1 = -5\sqrt3 v_0$[/tex]
[tex]$=-5\sqrt3 \left<\frac{1}{2},\frac{\sqrt3}{2} \right>$[/tex]
[tex]$= \left<-\frac{5 \sqrt3}{2},-\frac{15}{2} \right>$[/tex]
The unit vector in the direction normal to the plane, [tex]v_1[/tex] has components :
[tex]$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$[/tex]
[tex]$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$[/tex]
Therefore, [tex]$v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$[/tex]
From figure,
[tex]|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5[/tex]
∴ [tex]F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>[/tex]
[tex]$=\left<\frac{5 \sqrt3}{2},-\frac{5}{2} \right>$[/tex]
Therefore,
[tex]$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$[/tex]
[tex]$=<0,- 10> = F$[/tex]
