Take the 2-minute tour ×
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.

Let $P_1,P_2,P_3$ be $3$ different points in $\mathbb{R}$, then $P_1,P_2,P_3$ form a triangle.

What is the relation between the (one of the) angles of this triangle and $\langle P_2-P_1,P_3-P_1 \rangle$ ?

I am having trouble figuring out which angle of the triangle it is - and even why it is an angle in the triangle...

Note: $\langle ,\rangle$ denotes the standard inner product of $\mathbb{R}^2$.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

From the definition of the inner product for vectors $\vec{a},\vec{b}$ we got:

$$ \vec{a}\cdot\vec{b}=\|\vec{a}\| \|\vec{b}\|\cos(\widehat{\vec{a},\vec{b}}) $$

Applying this rule in your case:

$$ \langle P_2-P_1,P_3-P_1 \rangle =\|P_2-P_1\|\cdot \| P_3-P_1\|\cdot \cos(\widehat{P_2-P_1,P_3-P_1}), $$

so by reforming:

$$ \cos(\widehat{P_2-P_1,P_3-P_1})=\frac{\langle P_2-P_1,P_3-P_1\rangle}{\|P_2-P_1\|\cdot \| P_3-P_1\|} $$

The angle inside the $\cos$ is actually the angle between the two vectors that you're computing their inner product, in other words the angle between the two sides of your triangle formed by $P_2,P_1$ and $P_3,P_1$ respectively. So because the only mutual point is $P_1$, the angle is actually the angle corresponding to this point.

share|improve this answer

The angle between two vectors $a,b$ in $\mathbb{R}^2$ is actually defined by the formula $$ \cos(\varphi)=\frac{\langle a,b \rangle}{\Vert a\Vert\cdot\Vert{b}\Vert} $$ so $\frac{\langle P_2-P_1,P3-P_1\rangle}{\Vert P_2-P_1\Vert\cdot\Vert P_3-P_1\Vert}$ is the cosine of the angle in $P_1$.

Here's a motivation for the above formula: First note that we can rewrite $\frac{\langle a,b\rangle}{\Vert a\Vert\cdot\Vert{b}\Vert}=\left\langle \frac{a}{\Vert a\Vert},\frac{b}{\Vert b\Vert}\right\rangle $, so replacing $a$ and $b$ by $\frac{a}{\Vert a\Vert}$ and $\frac{b}{\Vert b\Vert}$ we can assume that the vectors have length $1$. Now write $b$ in the basis $a,a^{\perp}$ ($a^{\perp}$ is one of the 2 perpendicular vectors to $a$ of length $1$). In this basis $a$ corresponds to the vector $(1,0)$ and $b$ will be a vector on the unit circle, $b=x_1 a + x_2 a^{\perp}$. Note that the cosine of the angle between $a$ and $b$ is $x_1$ by the classical definition! This is in accordance to our formula $\cos(\varphi)=\langle a,x_1 a + x_2 a^{\perp}\rangle=\langle a,x_1 a\rangle =x_1$.

share|improve this answer

Your Answer

 
discard

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.