After thinking longing i can't figure it out no matter what..

So i have 3d line starts (0,0,0) and ends (3.5,3.5,2.5) so therefore has length of about 5.

Now how do i find out vector that is completely perpendicular to any vector like this one (1,2,3) assume all vectors start at 0,0,0.

If it was 2d i'd have used slope to find perpendicular line.

Please use more English and if you put letters like t or stuff explain it along too.

  • $\begingroup$ i know how to find opposite vector. opposite vector of (1,2,3) would be (-1,-2,-3) $\endgroup$
    – Techsin
    Sep 17, 2013 at 18:56
  • $\begingroup$ i dont know vector math $\endgroup$
    – Techsin
    Sep 17, 2013 at 18:57
  • 1
    $\begingroup$ I found this good pdf to explain it, if can help. $\endgroup$ Jul 5, 2014 at 17:08

3 Answers 3


The vectors orthogonal to $\mathbf{u} \in \mathbb{R}^n$ are the vectors $\mathbf{v} \in \mathbb{R}^n$ such that $\mathbf{u} \cdot \mathbf{v}=0$, where $\cdot$ is the dot product. In the case $\mathbf{u}=(3.5,3.5,2.5)$, we can find all such $\mathbf{v}=(a,b,c) \in \mathbb{R}^3$ by solving the equation defined by $$(3.5,3.5,2.5) \cdot (a,b,c)=0.$$ That is, all vectors $(a,b,c)$ such that $$3.5a+3.5b+2.5c=0$$ will be orthogonal to $(3.5,3.5,2.5)$.

  • $\begingroup$ i dont understand where did you get a,b,c...i have only one vector. And i want to find perpendicular vector. $\endgroup$
    – Techsin
    Sep 17, 2013 at 23:20
  • $\begingroup$ To begin with, $(a,b,c)$ represent the coordinates of an arbitrary vector. By assuming $(3.5,3.5,2.5) \cdot (a,b,c)=0$, we're insisting that $(3.5,3.5,2.5)$ and $(a,b,c)$ are orthogonal. Now any values of $a$, $b$ and $c$ that satisfy $3.5a+3.5b+2.5c=0$ will give an orthogonal (=perpendicular) vector $(a,b,c)$. Keep in mind there will be an infinite number of perpendicular vectors (in fact, there will be 2 free variables). $\endgroup$ Sep 17, 2013 at 23:32
  • $\begingroup$ ok i guess i am starting to understand better what i was asking...but still in confused a little i will come back with more..For now please consider this and 'get' what i am asking :D.. how do i find a vector that is perpendicular to another vector, and both vector start at 0,0,0. Now i can picture that all possibilities create a circle. So find the one that straight in parents' direction octants. I just realized vectors don't represent rotation. But rotations can be represented with vectors. $\endgroup$
    – Techsin
    Sep 17, 2013 at 23:50
  • $\begingroup$ I wish i could show an animation. suppose if vector is at (0,5,0) and has been rotated to 0deg(at xAxis) around Y-axis then perpendicular vector lies at (5,0,0). But how would i know this for other parent vectors $\endgroup$
    – Techsin
    Sep 17, 2013 at 23:54
  • $\begingroup$ If you want just one example in the $(3.5,3.5,2.5)$ case, we can just pick one solution, e.g. $(-1,1,0)$ will be perpendicular. So will $(-2,2,0)$ and infinitely many others. (In the $(0,5,0)$ case, any vector of the form $(a,0,b)$ will be perpendicular; so you can choose any random numbers for $a$ and $b$ here.) $\endgroup$ Sep 18, 2013 at 0:20

There exists a subspace of perpendicular vectors for any given vector. To find a perpendicular vector to any two vectors you can take their cross-product. To obtain a desired length, normalize and multiply by the desired length.

Consider the inner product: $\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta$. What conditions can you impose to derive unknown components?


Definition of the Dot Product:

$\vec{a} \cdot \vec{b}$ = ( $a_{1} , a_{2}$ ) $\cdot$ ( $b_{1} , b_{2}$ ) = $a_{1}b_{1} + a_{2}b_{2}$

also known as the scalar product or inner product

$\mathbf{\vec{a} \cdot \vec{b}}$ is a one "number" answer

Orthogonal Vectors:

Two vectors are orthogonal (perpendicular) if and only if $\ \mathbf{\vec{a} \cdot \vec{b} = 0}$ in other words... two vectors are perpendicular if their DOT PRODUCT is ZERO



$\vec{a}$ = ( 8 , -4 )

that is:

$a_{1}$ = 8

$a_{2}$ = -4

Find a vector $\mathbf{\vec{r}}$ that is perpendicular to $\mathbf{\vec{a}}$:

$\vec{r}$ = (x, y);

that is:

$b_{1} = x$

$b_{2} = y$

$\vec{a} \cdot \vec{r} = 8x + (-4y) = 0 \Rightarrow$

$\Rightarrow 8x - 4y = 0 \Rightarrow$

$8(1) - 4(2) = 0 \Rightarrow \mathbf{\vec{r} = (1, 2)} \Rightarrow$ one solution

$8(2) - 4(4) = 0 \Rightarrow \mathbf{\vec{r} = (2, 4)} \Rightarrow$ other solution

$8(-1) - 4(-2) = 0 \Rightarrow \mathbf{\vec{r} = (-1, -2)} \Rightarrow$ other solution

... so, I saw in this example what Rebecca said: << Keep in mind there will be an infinite number of perpendicular vectors >> ... and I shared it


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