chivistaelmore

2022-07-22

Let $\stackrel{\to }{a}=i+2j+3k$ and $\stackrel{\to }{b}=2i+5k$. For which value of t when $-2\le t\le 2$ holds the length of the vector $\stackrel{\to }{{c}_{t}}=t\stackrel{\to }{a}+\left(1-t\right)\stackrel{\to }{b}$ is as small as possible?
How should one approach this? The minimum would be when $\stackrel{\to }{{c}_{t}}$ is perpendicular to some vector $\stackrel{\to }{d}=\stackrel{\to }{b}-\stackrel{\to }{a}$ or is there something else Im not seeing?

Jazlene Dickson

Expert

Hint:
Write out the magnitude as a function of $t$ and then differentiate to find the maximum.

Pierre Holmes

Expert

Write $\stackrel{\to }{{c}_{t}}=t\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)+\stackrel{\to }{b}$. Then,
$\begin{array}{rl}|{\stackrel{\to }{c}}_{t}{|}^{2}& =|\stackrel{\to }{a}-\stackrel{\to }{b}{|}^{2}{t}^{2}+2\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)\cdot \stackrel{\to }{b}\phantom{\rule{mediummathspace}{0ex}}t+|\stackrel{\to }{b}{|}^{2}\\ & =|\stackrel{\to }{a}-\stackrel{\to }{b}{|}^{2}{\left(t+\frac{\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)\cdot \stackrel{\to }{b}}{|\stackrel{\to }{a}-\stackrel{\to }{b}{|}^{2}}\right)}^{2}-\frac{\left(\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)\cdot \stackrel{\to }{b}{\right)}^{2}}{|\stackrel{\to }{a}-\stackrel{\to }{b}{|}^{2}}+|\stackrel{\to }{b}{|}^{2}\\ & \ge \frac{\left(|\stackrel{\to }{a}-\stackrel{\to }{b}||\stackrel{\to }{b}|{\right)}^{2}-\left(\left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)\cdot \stackrel{\to }{b}{\right)}^{2}}{|\stackrel{\to }{a}-\stackrel{\to }{b}{|}^{2}}=|{\stackrel{\to }{c}}_{t}{|}_{min}^{2}\end{array}$

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