Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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

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

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

share|cite|improve this answer
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?

share|cite|improve this answer
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

share|cite|improve this answer
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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.