If $a$ and $b$ are vectors, such that $\|{a}\| = 7$ and $\|{b}\| = 11$, then find the smallest possible value of $\|{a} + b\|$.

Question

If $a$ and $b$ are vectors, such that $\|{a}\| = 7$ and $\|{b}\| = 11$, then find the smallest possible value of $\|{a} + b\|$.

Answer 1

Imagine vector $b$, which has length $11$ in some direction.   Adding to this is vector $a$, of length $7$ in some other direction.   Thus forming two sides of a triangle, the third side of length $\lVert a+b\rVert$.

What direction must vector $a$ point, relative to $b$, to make the size of $a+b$ the smallest?

Then the mininum plausible value of $\lVert a+b\rVert$ becomes obvious.

Answer 2

Let $\vec A + \vec B = \vec R$. We know that a = $|\vec A|$ = 7 and b = $|\vec B|$ = 11.

We have a formula $$|\vec R| = \sqrt{|\vec A|^{2} + |\vec B|^{2} + 2|\vec A||\vec B|\cos\theta}$$

The magnitude of $\vec R$ will be minimum if the angle between $\vec A$ and $\vec B$ is 180°.

We get $$|\vec R| = \sqrt{7^{2} + 11^{2} + 2(7)(11)\cos(180°)}$$ $$\therefore |\vec R| = \sqrt{49 + 121 - 154}$$ $$\therefore |\vec R| = \sqrt{16}$$ $$\therefore |\vec R| = 4$$

Answer 3

By the triangle inequality $$\|a+b\|+\|-a\|\geq\|a+b-a\|=\|b\|.$$ Thus, $$\|a+b\|\geq\|b\|-\|a\|=11-7=4.$$ The equality occurs, when $a=-\frac{\|a\|}{\|b\|}b,$ which says that $4$ is a minimal value.

Done!

Answer 4

Minimize $ || \vec a + \vec b || ^2$ :

$||\vec a + \vec b||^2 = ||a||^2 + ||b||^2 + 2 \vec a \cdot \vec b$.

The problem is reduced to minimizing the scalar product

$2\vec a \cdot \vec b = ||\vec a|| \cdot ||\vec b|| cos( \phi)$.

$-1 \le cos(\phi) \le 1$.

Choose $ \phi = π$.

$\vec a$ and $\vec b$ are antiparallel.

Minimum :

$||\vec a + \vec b|| ^2 =$

$||\vec a||^2 +||\vec b||^2 - 2||\vec a|| \cdot ||\vec b||$.

It remains to plug in the numbers.

And the maximum value?