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A particle is kept at rest at origin. Another particle starts from $(5,0)$ with a velocity of $-4i+3j$. Find the closest distance of approach.

Irregular User
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Dikshita
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    Welcome to MSE. People dislike helping you with homework unless you show some evidence of having tried to do the problem yourself. So what have you tried? What was the difficulty? More specifically, where do you think the particle will cross the $y$-axis? – almagest Jun 16 '16 at 14:27
  • Find the value of $t$ sothat the distance of the line $\vec{p}(t) = \binom{5}{0}+t\binom{-4}{3} $ to $(0,0)$ is minimized. (Why is this the equation for the line?). That is, minimize $|\vec{p}(t) - (0,0)|$. You can also minimize $|\vec{p}(t) - (0,0)|^2$ to get rid of the square-roots. – Maximilian Gerhardt Jun 16 '16 at 14:29
  • The slope of the line on which the particle is traveling is -3/4. You have a point on the line. Find the line equation and then use the standard formula to find the distance between the origin and this line. – Student Jun 16 '16 at 14:32
  • You should become familiar with the normal form of the equation of a line and how it allows problems like this to be quickly answered. – hardmath Jun 16 '16 at 16:04

3 Answers3

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The position over time is $(5,0)+(-4,3)t=(5-4t,3t)$, and the squared distance to the origin $(5-4t)^2+(3t)^2$.

Find the minimum value of this polynomial.

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The path meets the $y$-axis at some point $B$. The ratio $OB:OA$ is given as 3:4, so $OB=\frac{15}{4}$. By Pythagoras the ratio $OB:OA:AB$ is 3:4:5, so $AB=\frac{25}{4}$.

$D$ is the point of closest approach, so $\angle ODB=90^o$. Hence triangles $OBD,ABO$ are similar. Hence $\frac{OD}{OB}=\frac{OA}{AB}$ giving $OD=3$.

almagest
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The particle is at $$\vec{p}_0=\begin{pmatrix}5 \\ 0\end{pmatrix}$$ And has a directional velocity vector of $$-4\vec{i}+3\vec{j} = \binom{-4}{3}$$ Therefore, the line this particle is traveling on is simply $$\vec{p}(t) = \binom{5}{0}+t\binom{-4}{3}$$ Or, as a line equation $y=ax+b$: We know through the velocity vector that the slope is $$a = \frac{\Delta y}{\Delta x} = \frac{3}{-4} = -\frac{3}{4}.$$

We know that $(5,0)$ is on the line, so: $$0 =-\frac{3}{4}\cdot 5+b$$ $$0 =-\frac{15}{4}+b$$ $$\frac{15}{4} =b$$ So we must have $$y=-\frac{3}{4}x+\frac{15}{4}$$ graph

We setup a function that measures how far away the a point on the line $p(t)$ is away from the origin $(0,0)$. That is, we simply compute the vector-distance between $\vec{p}(t)$ and $\vec{0} =\binom{0}{0}$ as: $$d(t) = \left\vert\vec{p}(t) - \vec{0} \right\vert = |\vec{p}(t)| = |(5-4t)\vec{i} + (0+3t)\vec{j} |= \sqrt{(5-4t)^2+(3t)^2} $$

The distance has to be minimal. So, this is an optimization problem on finding the value of $t$ for which $d(t)$ will be minimal. But, we can also minimize the squared-distance, that is $$d(t)^2= (5-4t)^2+(3t)^2$$ We expand. $$d(t)^2= 25t^2-40t+25$$ We compute the derivative. $$(d^2(t))'= 50t-40$$ We find the zeroes. $$50t-40=0$$ $$50t=40$$ $$t=\frac{40}{50} = \frac{4}{5} = 0.8$$

We plug this value into $p(t)$ to find the point on the line:

$$\vec{p}(0.8) = \binom{1.8}{2.4}$$ And the distance minimal distance to $(0,0)$ is $$\left\vert \binom{1.8}{2.4}\right\vert = \sqrt{1.8^2 + 2.4^4} = \sqrt{9} = 3$$ Which is the final answer, the minimal distance of approach is equal to $3$.

Maximilian Gerhardt
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