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This is a simple problem I am having a bit of trouble with. I am not sure where this leads.

Given that $\vec a = \begin{pmatrix}4\\-3\end{pmatrix}$ and $|\vec b|$ = 3, determine the limits between which $|\vec a + \vec b|$ must lie.

Let, $\vec b = \begin{pmatrix}\lambda\\\mu\end{pmatrix}$, such that $\lambda^2 + \mu^2 = 9$


$$ \begin{align} \vec a + \vec b &= \begin{pmatrix}4+\lambda\\-3 + \mu\end{pmatrix}\\ |\vec a + \vec b| &= \sqrt{(4+\lambda)^2 + (\mu - 3)^2}\\ &= \sqrt{\lambda^2 + \mu^2 + 8\lambda - 6\mu + 25}\\ &= \sqrt{8\lambda - 6\mu + 34} \end{align} $$

Then I assumed $8\lambda - 6\mu + 34 \ge 0$. This is as far I have gotten. I tried solving the inequality, but it doesn't have any real roots? Can you guys give me a hint? Thanks.

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In general, $$\Bigl| \lVert \mathbf{v}\rVert - \lVert \mathbf{w}\rVert\Bigr| \leq \lVert \mathbf{v}\pm\mathbf{w}\rVert \leq \lVert\mathbf{v}\rVert + \lVert\mathbf{w}\rVert.$$Your assumption is unneeded: that expression equals a sum of two squares, so it must be nonnegative. –  Arturo Magidin Jul 27 '11 at 5:38
Thanks, need to practice some triangle inequality problems. –  mathguy80 Jul 27 '11 at 5:54

5 Answers 5

up vote 3 down vote accepted

We know that $\|a\|=5$, $\|b\|=3$, and we have two vector formulas

$$ \|a+b\|^2=\|a\|^2+2(a\cdot b)+\|b\|^2,$$

$$ a\cdot b = \|a\| \|b\| \cos\theta.$$

Combining all this, we have

$$\|a+b\|^2 = (5^2+3^2)+2(5)(3)\cos\theta.$$

Cosine's maximum and minimum values are $+1$ and$-1$, so we have

$$\|a+b\|^2 \in [4,64]$$

$$\|a+b\| \in [2,8].$$

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Very neat, Thank you! –  mathguy80 Jul 27 '11 at 5:49

So you've got a circle of radius 3 centered on $(4,-3)$ and you want to find the points nearest and farthest from the origin. Draw a line through the origin and the center of the circle, meeting the circle at two points. Can you see why those two points are the ones you're looking for?

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Excellent! I wasn't thinking in terms of graphs. Thank you. –  mathguy80 Jul 27 '11 at 5:52

Hint 1: Triangle inequality

Hint 2: Reverse triangle inequality

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Good hint, I hadn't thought of using triangle inequality in terms vectors. Thanks. –  mathguy80 Jul 27 '11 at 5:50

Given $λ^2+μ^2=9$ (1), you need to find the maximum and minimum of y = $8*\lambda - 6* \mu$ (2)

One way to do it, I think is to substitute $\lambda$ from (2) into (1), get a quadratic equation of $\mu$, with the parameter of y. You need this equation to have a root, from that you could find the range values of y

Another way is to use graph. In particular, you draw graph of (1), you get a circle, and the graph of (2): $\lambda = y/8 + 3/4 \mu$. With various values of y, you will get various lines parallel to $\lambda = 3/4 \mu$, and you want the min and max of y, so that the line will still cut the circle (the line intersects the $O\lambda$ at y/8)

A third way, more general way(in Calculus 3) is to use the Lagrange Multiplier method

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Nice answer, answered my question and also gave me pointers on what I need to study further. Thanks. –  mathguy80 Jul 27 '11 at 5:53

I would parameterize $b$ as $(3 \cos t, 3 \sin t)$, so the points to look at are $(4 + 3 \cos t, -3+ 3 \sin t)$. Find the value of $t$ that gives extreme values for the distance of this from the origin.

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