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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?

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    $\begingroup$ 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. $\endgroup$
    – David K
    Commented Jul 30, 2019 at 3:49
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    $\begingroup$ 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 $\endgroup$ Commented Jul 30, 2019 at 3:58

3 Answers 3

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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.

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  • $\begingroup$ can you please repost your answer to my question here? I really like your well presented answer. Thanks! $\endgroup$
    – J. Doe
    Commented Jan 14, 2020 at 7:36
  • $\begingroup$ @J.Doe Done. I deleted it because I thought it merely covered the same ground as the answer that I found already there when I posted it. $\endgroup$ Commented Jan 14, 2020 at 7:47
  • $\begingroup$ Thank you!!! I really appreciate it. $\endgroup$
    – J. Doe
    Commented Jan 14, 2020 at 7:48
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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.

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    $\begingroup$ Thank you! I was able to find the solution immediately after your and Lonza's answer. $\endgroup$
    – Liam
    Commented Jul 30, 2019 at 3:58
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Hint:

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.

enter image description here

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