Why are vectors  and <-b,a> perpendicular? If i have the vectors:
X = {a, b}
Y = {-b, a}
How can I explain that these vectors will always be perpendicular? I know I can prove this very easily via the dot product, but I need to explain it in a layman's way. 
 A: Let me assume $a, b \geq 0$
If you rotate a vertical line that points upward by $90^\circ$ counter clockwise, it becomes a horizontal line that point to the left.
That is $\begin{bmatrix} 0 \\ b\end{bmatrix}$ will be rotated to become $\begin{bmatrix} -b \\ 0\end{bmatrix}$
If you rotate a horizontal line that points to the right by $90^\circ$ counter clockwise , it becomes a vertical line that point upward.
That is $\begin{bmatrix} a \\ 0\end{bmatrix}$ will be rotated to become $\begin{bmatrix} 0 \\ a\end{bmatrix}$
In summary , if you rotate $\begin{bmatrix} a \\ b\end{bmatrix}=\begin{bmatrix} a \\ 0\end{bmatrix}+\begin{bmatrix} 0 \\ b\end{bmatrix} 90^\circ$  counter clockwise, you obtain $\begin{bmatrix} 0 \\ a\end{bmatrix}+\begin{bmatrix} -b \\ 0\end{bmatrix}=\begin{bmatrix} -b \\ a\end{bmatrix}$. You might want to draw a horizontal line segment that start from $(0,0)$ to $(0,a)$ and a vertical line segment joining $(0,a)$ to $(a,b)$ and rotate it.
Hence $\begin{bmatrix} a \\ b\end{bmatrix}$ and $\begin{bmatrix} -b \\ a\end{bmatrix}$ are perpendicular.
A: Option 1: Angular Algebra
$(a,b)$ has $\theta$ degrees with $x$.
$(b,a)$ flips $x$ by $y$, with $90-\theta$.
$(-b,a)$ mirrors with $y$, with $180-90+\theta=90+\theta$.
Then $(a,b)$ being perpendicular to $(-b,a)$.
Option 2: Geometry
$OAX$ is the rectangular triangle with the origin, the point A (a,b) and the point X (a,0) in the x axis.
Rotate this triangle in 90 degrees, keeping the origin, changing the point (a,0) to (0,a), and turning the point A into (-x,a). The value of x is trivially b.
Option 3: Polar Vectors
Let be $(a,b)=r(\cos\theta,\sin\theta)$. Constrain $r$ and increase freely in $\theta$.
Get the orbit derivative to obtain the tangent derivative $r(-\sin\theta,\cos\theta)$. They are clearly perpendicular.
A: Draw the vector $X$ on a grid paper. To do this, you count $a$ squares to the right, $b$ squares up.
Then rotate the whole paper 90 degrees counterclockwise.
Behold $Y$!
(Namely: $a$ squares up, and $b$ squares to the left = $-b$ squares to the right.)
(For simplicity you can assume that $a$ and $b$ are positive integers to begin with.
Then convince yourself that the same thing works even when $a$ and $b$ are not positive or integers.)
A: I think the best way is by graphing the two for some particular values of $(a,b)$. Then, you can show how the angle between $(a,b)$ and the $x$-axis is equal to the angle between $(-b,a)$ and the $y$-axis (congruent triangles), so the angle between $(a,b)$ and $(-b,a)$ is $90^{\mathrm{o}}$. 
A: $$ \begin{pmatrix} -b \\ a \end{pmatrix} = R_{90^o}\begin{pmatrix} a \\ b \end{pmatrix}= \begin{pmatrix} \cos( \pi/2) & -\sin(\pi/2) \\ \sin(\pi/2) & \cos(\pi/2) \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix}$$
A: A bit Geometry:
Let $\vec A := (a,b)$;  $\vec B := (-b,a)$.
Draw the vectors from the origin $ (0,0)$;
label the endpoints $A$ and $B$ respectively.
Vector addition: 
$\vec C: = \vec B - \vec A = $($-b - a$,  $a - b$);
$ \triangle OAB$:
$|\vec A|^2 = a^2 + b^2$;
$|\vec B|^2 = a^2 + b^2$;
$|\vec C|^2 = (a +b)^2 + (a- b)^2 =$
$2a^2 + 2b^2$.
The lengths of the 3 vectors are the lengths of the 3 sides of $\triangle OAB$.
Pythagoras:
$|\vec A|^2 + |\vec B|^2 = |\vec C|^2$;
$\triangle OAB$ is a right triangle with the right angle at $O$,
$\vec A$ is perpendicular to $\vec B$.
Alternative:
Consider the 2 straight lines:
1) Along vector $\vec A$:  $y = m_1x$, where $m_1 = (b/a)$;
2) Along vector $\vec B$ :  $y = m_2x$, where $m_2 = (a/-b)$.
$m_1 m_2 = - 1$.
The 2 lines are perpendicular.
A: One way: Show that the triangle built on them satisfies the Pythagorean theorem, which implies that it's a right triangle
Another way: Show that the four points $(a,b)$, $(-b,a)$, $(-a,-b)$, and $(b,-a)$ form a rhombus, each of whose vertices are equidistant from the center. This implies that they form a square.
A third way: Form the right triangle with vertices $(0,0)$, $(0,b)$, and $(a,b)$. In addition, form the right triangle with vertices $(0,0)$, $(0,a)$, and $(-b,a)$. Observe that these are congruent, and fill in the angles.
A: You can argue that $(a,b)\mapsto(-b,a)$ is a $90^°$ rotation be applying this operation twice:
$$(a,b)\quad\mapsto\quad(-b,a)\quad \mapsto\quad(-a,-b).$$
The latter one is obviously the $180^°$ rotated vector to $(a,b)$. As you reached at this point by doing the former operation twice it seems as this would be the half of a $180^°$ turn, and this is a $90^°$ turn. I hope it is layman-terms that such a rotation will make a perpendicular vector.
