# Range of $|a+b|$ given $a = ⟨15,-8⟩$ and $|b| = 12$

The question follows:

Given vectors $$a,b$$ such that $$a = ⟨15, -8⟩$$ and that $$|b| = 12$$, find the range of possible values of $$|a+b|$$.

I have been attempting to solve this question for the past 3 hours:

Let $$b = ⟨x,y⟩$$. I believe that the answer lies in the fact that $$|b| = 12 = \sqrt{144} = \sqrt{x^2 + y^2}$$. The question states that I need to find the range of possible values, i.e. the maximum and minimum possible values.

What I have also worked out is $$|a+b| = |⟨15 + x, -8 + y⟩|$$. Now I think I need to determine the values of $$x$$ and $$y$$ such that I obtain the maximum and minimum values of $$|a+b|$$, such that the root of the sum of the squares of $$x$$ and $$y$$ still adds up to $$12$$.

Unfortunately, I do not know how to do this, hence, could someone nudge me in the right direction or show me how?

• After looking at the answers below, your next step should be to figure out the value of $|a|.$ You don't need to write $b$ as $x$ and $y$ coordinates. Commented Jul 30, 2019 at 3:49
• Welcome to Math Stack Exchange. Draw a circle of radius $12$ with center $(15,-8)$. The points on the circle closest to and farthest from $(0,0)$ will be on the line through $(0,0)$ and $(15,-8)$. You can figure out the distance from $(0,0)$ to $(15,-8)$ and add or subtract the radius of the circle to get the distance from $(0,0)$ to those points on the circle Commented Jul 30, 2019 at 3:58

Hint: $$\left\vert\mathbf{a}+\mathbf{b}\right\vert^2=\left\vert\mathbf{a}\right\vert^2+\left\vert\mathbf{b}\right\vert^2+2\,\mathbf{a}\cdot\mathbf{b}\$$, and, for fixed $$\ \mathbf{a}\$$ and $$\ \left\vert\mathbf{b}\right\vert\$$, $$\ \mathbf{a}\cdot\mathbf{b}\$$, is a maximised when $$\ \mathbf{b}\$$ has the same direction as $$\ \mathbf{a}\$$, and minimised when it has the opposite direction.

First, convince yourself that if $$x$$ and $$y$$ are vectors, then $$|x|-|y|\le|x+y|\le|x|+|y|$$. Further convince yourself that the left inequality is an equality if and only if $$x$$ and $$y$$ are anti-parallel (that is, point in opposite directions), while the right inequality is an equality if and only if $$x$$ and $$y$$ point in the same direction. Then you should have no difficulty answering your question, and lots of others, besides.
Shown (in green) is a circle of radius $$12$$ centered at point $$(15,-8).$$ The points on the circle closest to and farthest from $$(0,0)$$ will be on the line (shown in blue) through $$(0,0)$$ and $$(15,−8)$$. Figure out the distance from $$(0,0)$$ to $$(15,−8)$$, and add or subtract the radius of the circle to get the distance from $$(0,0)$$ to those points on the circle.