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Let's take a classic 2-dimensiona vector addition:

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Given vector $ \vec{a} $ and angle $\phi$ I need to find $ \vec{b} $ of least possible magnitude such that vector $ \vec{c} = \vec{a}+ \vec{b} $ is at angle $\phi$ to $ \vec{a} $; I don't care about magnitude of $ \vec{c} $.

(actual practical problem I face, optimal angle of burn of a rocket engine to change rocket trajectory by given angle at moment's notice in Kerbal Space Program. I found how to retain magnitude $ | \vec{c} | = |\vec{a}| $ but I'm really not sure how to approach optimizing it when the resultant magnitude is insignificant and I just need to change direction for least cost, simply accepting whatever speed results from that.)

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    Hint: $b\perp c$ – 2017-01-26
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    $b=|\sin \theta| Ra$, where $R$ is a rotation of $\pi-\theta$. – 2017-01-26
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    This is equivalent to finding the closest point to $c$ on the line defined by $a$. – 2017-01-26
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    @Exodd: Actually your hint solves my practical problem. Point my craft perpendicular to where I want to go, burn until direction is right. – 2017-01-26

2 Answers 2

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We know that $$ \|b\|^2 = \|c-a\|^2 = \|a\|^2 + \|c\|^2 - 2\|a\|\|c\|\cos \phi $$ $\|a\|$ and $\theta$ are fixed, so we can only change $\|c\|$. We can find the minimum of $\|b\|$ by deriving in $\|c\|$ and putting the derivative to zero: $$ 2\|c\|-2\|a\|\cos \phi = 0 $$ $$ \|c\| = \|a\| \cos \phi $$ $$ \|b\|^2 = \|a\|^2\sin^2\phi $$ so you obtain that $b\perp c$ since $$ \|a\|^2 = \|c-b\|^2 = \|c\|^2 + \|b\|^2 - 2 c\cdot b = \|a\|^2 - 2c\cdot b $$ $$\implies c\cdot b =0$$

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    While Ross's answer is more brief and looks quite elegant, your is easier to use in practical computation. (btw, it may seem obvious, but besides $b\perp c$, to make the answer complete, $ \|b\| = \|a\|\sin\phi $ ). – 2017-01-27
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You want to project $\vec a$ onto the direction $\vec c$. That projection is $(\vec a \cdot \vec c)\vec c$. Then $\vec b = \vec a -(\vec a \cdot \vec c)\vec c$