2. ## Re: interception

Vdc =(15cosa-15)i+(15sina-15√3)j
if you look at skeeter diagram this is this represents the line labelled △r
$V_{dc}$ is the relative velocity vector, $\color{red}{V_r}$ ... the relative motion (position) vector, $\color{red}{\Delta r}$ is parallel to $\color{red}{V_r}$.

$\cos{\theta} + \sqrt{3}\sin{\theta} = 1$

$\sqrt{3}\sin{\theta} = 1-\cos{\theta}$

$3\sin^2{\theta} = 1 - 2\cos{\theta} + \cos^2{\theta}$

$3(1-\cos^2{\theta}) = 1 - 2\cos{\theta} + \cos^2{\theta}$

$0 = 2\cos^2{\theta} - \cos{\theta} - 1$

$0 = (2\cos{\theta} + 1)(\cos{\theta} - 1) = 0$

$\cos{\theta} = -\dfrac{1}{2} \implies \theta = 120^\circ$

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