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Points $A$ and $B$ have position vectors $\vec{OA} = \begin{pmatrix}2\\ 2\\ 3\end{pmatrix}$ and $\vec{OB} =\begin{pmatrix}-1\\ 7\\ 2\end{pmatrix}$. Find the angle between $\vec{AB}$ and $\vec{OA}$.

So I found $\vec{AB}=\begin{pmatrix}-3\\ 5\\ -1\end{pmatrix}$.

Then to find the angle, shouldn't I change the sign of the components of the OA vector and then use the formula $\cos \left(\theta \right)=\frac{a\cdot b}{\left|a\right|\left|b\right|}$ ? My textbook mentions that vectors should be pointing away from the angle you are trying to find.

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So we have $AB=(-3,5,-1)$ and $OA=(2,2,3)$. Now use, for example, the formula

$$\frac{a\cdot b}{|a||b|}=\cos\theta,$$ where $\theta$ is the angle between $a$ and $b$.

Edit: I think the confusion here is what we really mean by angle between two vectors. Since vectors "float" in space, we can imagine placing them so that their tails touch. Then the angle between these two physical interpretations of vectors is defined as the angle between $a$ and $b$ as vectors. There is no ambiguity about where the vector is pointing because we place them tail-to-tail so both arrows always point away.

Here is a pretty bad picture I made in photoshop. Even though b "points away" from the angle, it doesn't really affect anything. enter image description here

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    $\begingroup$ Yeah, but don't the vectors have to be pointing away from the angle? Isn't OA pointed towards point A, where the angle is? $\endgroup$ – Chx Aug 24 '18 at 20:22
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    $\begingroup$ That doesn't make sense. The vector is not actually located anywhere in the plane. If the tail of your vector is at $A$ then $OA$ points away. We usually just define the angle between two vectors to be the smallest (i.e., between $0$ and $\pi$). $\endgroup$ – Elliot G Aug 24 '18 at 20:23
  • $\begingroup$ Sorry, but what do you mean by the vector is not located in the plane? $\endgroup$ – Chx Aug 24 '18 at 20:34
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    $\begingroup$ A vector in 3-space is nothing more than the three numbers $(x_1,x_2,x_3)$. We describe them as arrows because it helps with intuition. For example, the vector $(1,1)$ can be interpreted "go northeast by $\sqrt 2$." This is not the same thing as saying "start at the origin and go northeast by $\sqrt 2$." So with the arrow interpretation they float around in space. $\endgroup$ – Elliot G Aug 24 '18 at 20:36
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    $\begingroup$ When the vectors describe physical shapes, then their starting and ending place does matter. In my picture, the angle between $a$ and $b$ is greater than $90^\circ$ as line segments but less than $90^\circ$ as vectors. They differ by $180^\circ$ in this case, which is why your answer is off by $180^\circ$. ($-105.5+180=74.5$) $\endgroup$ – Elliot G Aug 24 '18 at 21:12
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You can ignore the direction of the vector. When you solve $\cos \theta = $junk, you have your choice of many values. Choose the value between $0$ and $\pi/2$ to get the acute angle between the vectors. That's it.

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  • $\begingroup$ okay. So when do I pay attention to the direction of the vector? I wonder why my textbook mentioned it. Also, in a previous question, I answered incorrectly because the sign on my vector was wrong. Thanks for replying! $\endgroup$ – Chx Aug 24 '18 at 20:30

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