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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 $\lambda^2 + \mu^2 = 9$

Then,

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

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.

mathguy80
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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 05:38
  • Thanks, need to practice some triangle inequality problems. – mathguy80 Jul 27 '11 at 05:54

5 Answers5

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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?

Gerry Myerson
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Hint 1: Triangle inequality

Hint 2: Reverse triangle inequality

JavaMan
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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

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

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

anon
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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

Long
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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.

marty cohen
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