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.

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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!

  • 1
    $\begingroup$ The vectors $\vec{AB}$ and $\vec{CD}$ are perpendicular, so the inner product of those two vectors should be zero. $\endgroup$ Jul 29, 2014 at 14:23
  • $\begingroup$ I'm not exactly sure what you are asking, you give the coordinates of $A,B,C,D$ and want to find $D$. Did you mean to use the symbol $D$ for different things? $\endgroup$
    – copper.hat
    Jul 29, 2014 at 14:28
  • $\begingroup$ I should have named the bold vector CD to CE. Sorry for the misunderstanding, my bad. $\endgroup$
    – axel
    Jul 29, 2014 at 14:36

2 Answers 2


You know two things:

  1. The dashed line $\vec{CD}$ is perpendicular to $\vec{AB}$
  2. 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.

  • $\begingroup$ Sorry but I cannot understand how to make the equation. Could you be more a little bit more descriptive. $\endgroup$
    – axel
    Jul 30, 2014 at 6:03
  • $\begingroup$ Are you familiar with inner products? $\endgroup$
    – MRicci
    Jul 30, 2014 at 6:30
  • $\begingroup$ Actually I am not, but I tried to learn from wikipedia without much success. $\endgroup$
    – axel
    Jul 30, 2014 at 7:09
  • $\begingroup$ In Euclidean geometry (on a plane or space in Cartesian coordinates), the inner product between two vectors is also known as the dot product, and is just the sum of the product of their coordinates: $A = (a_x,a_y)$ and $B = (b_x,b_y)$, then $A \cdot B = a_x b_x + a_y b_y$. When this is zero, the two vectors are perpendicular. (In this case, the point $A = (a_x,a_y)$ represents a vector starting at the origin and ending at $(a_x,a_y)$). $\endgroup$
    – LucasVB
    Jul 30, 2014 at 14:18

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.


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