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A particle at position $p$ and velocity $\vec{v}=\langle x,y,z \rangle$ hits the plane orthogonal to vector $\vec{n}$ and passing through point $q$. When does the particle hit the plane?

I calculated the distance between particle and plane, but my problem is how to convert vector into units so that I can calculate the time. I tried the magnitude, but it gives incorrect solution. Can someone point me in the right direction?

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I think we're supposed to assume the particle travels at constant velocity $\bf v$, and at time $t = 0$ has position ${\bf r}\left(0\right) = \bf p$, so that it's position as a function of time is $$ {\bf r}\left(t\right) = {\bf v} t + {\bf p}. $$ This vector lies in the plane if $$ \left[{\bf r}\left(t\right) - {\bf q}\right] \bullet {\bf n} = \left( {\bf v} \bullet {\bf n}\right) t + \left({\bf p} - {\bf q}\right) \bullet {\bf n} = 0 \Rightarrow t = \frac{\left({\bf p} - {\bf q}\right) \bullet {\bf n}}{{\bf v} \bullet {\bf n}}. $$

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what if the particle is big and has radius r, then t= ((p+r)-q).n/v.n ? where r is constant. Am I correct? – nullPointer2 Sep 8 '12 at 2:52
@BadSniper: No, there are two problems. First, you can't subtract a scalar (r) from a vector ($\bf p$). Second, you have shortened the vector $p$ by $r$, but you will hit the plane when the perpendicular is of length $r$. You would want to shorten $\bf p$ by $\frac r{\cos \theta}$ where $\theta$ is the angle between $\bf p$ and the plane. – Ross Millikan Sep 8 '12 at 3:06
p is not the vector. It's position. So I have to subtract coordinates by r/cosθ? – nullPointer2 Sep 8 '12 at 3:17

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