How to find perpendicular point of a vector to another vector 2d Given the axis x-y and some random points to the vectors AB and CD, how can i find out where will the point D lie when the vector CD(dashed line) is perpendicular to AB. For example if point A has coordinates (2,1), B (10,7), C (6,3), D (6,14), what will the coordinates of D be, if the vector CD is perpendicular to the vector AB.
 
I am not looking for a straight solution, but for some guidelines on how can I achieve that, since I am not very good in geometry. I guess it can be managed by using the angle between points of the vector AB, but I am not sure.
Thank you in advance!
 A: You know two things:


*

*The dashed line $\vec{CD}$ is perpendicular to $\vec{AB}$

*The dashed $\vec{CD}$ and the bold $\vec{CD}$ have the same length. Actually, just call the bold one something else, like $\vec{CE}$. 


The first condition means that $<B-A,C-D> = 0$. The second condition means that $\|C-E\| = \|C-D\|$. You have two unknowns, namely the coordinates of $D$. You can use these two conditions to solve for these unknowns. 
Note on inner products: You can think of an inner product as a function that captures the notion of angle. In your case, we can use the traditional dot product, a type of inner product. In the plane, for example, the dot product of $A = (a_1,a_2)$ and $B=(b_1,b_2)$ is written $A\cdot B$ and is given by \begin{equation} a_1b_1 + a_2b_2. \end{equation} A very important fact is that when two vectors are perpendicular (orthogonal), they have dot product equal to $0$. 
So, let the coordinates of $D$ be $(x,y)$. Expand the first condition using the formula for the dot product I gave above. Expand the second condition using the definition of the Euclidean norm (in this case, the familiar distance formula). This will give you a system of two equations in two unknowns. Solve it for $x$ and $y$. 
Note: Once you use these conditions to write a system of equations, you may want to think about how many solutions this system has and what this means geometrically. 
A: A quick way to think about it is: if you have a vector $(x,y)$, then rotating it counterclockwise $90^o$ you get $(-y,x)$. We have $\mathbf{AB} = (8,6)$. If $\mathbf{D} = (x_d,y_d)$, if you want $\mathbf{AB} \perp \mathbf{CD}$, then $\mathbf{CD} = (x_d - 6, y_d - 3)$ is parallel to $(-6,8)$. Hence exists $\lambda \in \Bbb R$ such that: $$(x_d - 6, y_d - 3) = \lambda (-6,8)$$
You have two equations and three unkowns, because you still have one degree of freedom: $\mathbf{CD}$'s lenght. 
