Minimum Vector Norm
If $ \mathbf{a} $ and $ \mathbf{b} $ are vectors such that $ \|\mathbf{a}\| = 3 $ and $ \|\mathbf{b}\| = 14 $, then find the smallest possible value of $ \|\mathbf{a} + \mathbf{b}\| $.
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- $\frac{a}{b}$
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Solution
We have that
\begin{align*}
\|a + b\|^2 &= (a + b) \cdot (a + b) \\
&= a \cdot a + 2 a \cdot b + b \cdot b \\
&= \|a\|^2 + 2 a \cdot b + \|b\|^2.\end{align*}We know that $ \|a\| = 3 $ and $ \|b\| = 14 $. Also, if $ \theta $ is the angle between the vectors $ a $ and $ b $, then
\[a \cdot b = \|a\| \cdot \|b\| \cos \theta = 42 \cos \theta.\]Hence,
\[\|a + b\|^2 = 205 + 84 \cos \theta.\]This quantity is minimized when $ \cos \theta = -1 $ (or $ \theta = 180^\circ $), which gives us
\[\|a + b\|^2 = 205 - 84 = 121,\]so the minimum value of $ \|a + b\| $ is $ \sqrt{121} = \boxed{11} $. (We have effectively proved the Triangle Inequality for vectors in this problem.)
Freeflows