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Let $a=(x_1,x_2,x_3)$ and $b=(y_1,y_2,y_3)$ be Euclidean vectors.

These vectors are said to be orthogonal iff $\displaystyle \sum_{i=1}^n x_iy_i=0$.

I have a hard time understanding how this formalisation captures the intuitive notion of orthogonality. How does it?

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    Are you familiar with scalar product? – 2017-02-16

4 Answers 4

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We proceed to develop an intuitive understanding of the inner product by appealing to a geometric picture. We then proceed to apply straightforward vector algebra.

First, we draw two vectors $\vec A$ and $\vec B$ that originate at the origin. The vector $\vec A-\vec B$ is the vector from $\vec B$ to $\vec A$. Therefore, the three vectors $\vec A$, $\vec B$, and $\vec A-\vec B$ form a triangle.

From the Law of Cosines, we have

$$|\vec A|^2+|\vec B|^2-2|\vec A|\,|\vec B|\cos(\theta)=|\vec A-\vec B|^2 \tag 1$$

where $\theta\in [0,\pi]$ is the angle between $\vec A$ and $\vec B$.

But note that

$$\begin{align} |\vec A-\vec B|^2=&(A_1-B_1)^2+(A_2-B_2)^2+(A_3-B_3)^2\\\\ &=\color{red}{(A_1^2+A_2^2+A_3^2)}+\color{green}{(B_1^2+B_2^2+B_3^2)}-2\color{blue}{(A_1B_1+A_2B_2+A_3B_3)}\\\\ &=\color{red}{|\vec A|^2}+\color{green}{|\vec B|^2}-2\color{blue}{(\vec A\cdot \vec B)}\tag2 \end{align}$$

Using $(1)$ in $(2)$ yields

$$\vec A\cdot \vec B=|\vec A|\,|\vec B|\cos(\theta)\tag 3$$

Finally, note from $(3)$ that if $\vec A\cdot \vec B=0$, then $\cos(\theta)=0$, so that $\theta =\pi/2$, and thus $\vec A$ and $\vec B$ are orthogonal.

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The crucial insight comes (without any trigonometry) from the plane. Recall that two lines are orthogonal if and only if their slopes are negative reciprocals. (You can see this, if you want, by considering the right triangles formed by the vectors $\mathbf x=(a,b)$ and $\mathbf y=(-b,a)$. As you can see, the right angle makes the triangles congruent.

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But then we observe that $\mathbf x\cdot\mathbf y = a(-b)+b(a)=0.$

Algebraically, we can deduce that dot product $0$ implies perpendicularity. Algebraic properties of the dot product tell us that $\|\mathbf x+\mathbf y\|^2 = \|\mathbf x\|^2 + \mathbf y\|^2 \iff \mathbf x\cdot\mathbf y=0$. The converse of the Pythagorean Theorem will then tell us that this means $\mathbf x$ and $\mathbf y$ are perpendicular (orthogonal).

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    Just curious as to your thoughts Ted. So, please don't take this line of questioning as offensive - not my intent. So, how does one know, a priori, that two lines are perpendicular if and only if there slopes are negative reciprocals? Given the definition of orthogonality and algebraic structure of the inner product, it is clear. But how does one intuit perpendicularity without appealing to the physical world? And certainly, the Pythagorean Theorem is indeed trigonometric, is it not?. – 2017-02-17
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    @Dr.MV: Perhaps you didn't take the time to read what I wrote. I gave the argument that perpendicular implies relation between slopes, and the converse follows immediately, as well. I learned the geometry well before I saw any trigonometry. – 2017-02-17
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    Actually Ted, I did take the time to read what you wrote. I evidently lack the intuition needed to see that, while the inner product of the vectors that span two lines whose slopes are negative reciprocal is zero, the vectors are perpendicular without appealing to geometry/trigonometry. Perhaps I unfairly link geometry and trigonometry together. – 2017-02-17
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The dot product $$ \displaystyle \sum_{i=1}^n x_iy_i $$ can be shown to be equal to $$ \| x\| \| y\| \cos\theta $$ where $\theta$ is the angle between $x$ and $y$. Then it becomes clear that if the dot product is $0$ then $\theta = \pi/2$, which means, geometrically, that $x$ and $y$ are orthogonal (perpendicular).

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You have used $$\vec{a} \cdot \vec{b} = \sum_{i=1}^n x_i y_i$$ but $$\vec{a} \cdot \vec{b} = \left\| \vec{a} \right\| \left\|\vec{b} \right\| \cos \theta$$ where $\theta$ is the angles between $\vec{a}$ and $\vec{b}$, so $\vec{a} \cdot \vec{b} = 0$ either when $\vec{a} = \vec{0}$, or $\vec{b} = \vec{0}$ or $\theta = \pi/2 \pm n\pi$...