How to find perpendicular vector to another vector?

Question

How do I find a vector perpendicular to a vector like this: $$3\mathbf{i}+4\mathbf{j}-2\mathbf{k}?$$ Could anyone explain this to me, please?

I have a solution to this when I have $3\mathbf{i}+4\mathbf{j}$, but could not solve if I have $3$ components...

When I googled, I saw the direct solution but did not find a process or method to follow. Kindly let me know the way to do it. Thanks.

Answer 1

There exist infinite vector in 3 dimension that are perpendicular to a fixed one. They should only satisfy the following formula: $$(3\mathbf{i}+4\mathbf{j}-2\mathbf{k}) \cdot v=0$$

For finding all of them, just choose 2 perpendicular vectors, like $v_1=(4\mathbf{i}-3\mathbf{j})$ and $v_2=(2\mathbf{i}+3\mathbf{k})$ and any linear combination of them is also perpendicular to the original vector: $$v=((4a+2b)\mathbf{i}-3a\mathbf{j}+3b\mathbf{k}) \hspace{10 mm} a,b \in \mathbb{R}$$

Answer 2

You just need to find any vector $v \neq 0$ such that $v \cdot (3\mathbf{i}+4\mathbf{j}-2\mathbf{k}) = 0$.

There is no unique solution, any one will do. To save typing, let $p = 3\mathbf{i}+4\mathbf{j}-2\mathbf{k}$.

Pick a vector $x$, that is not on the line through the origin and $p$. Take $x = 3\mathbf{i}$, for example.

Construct a vector perpendicular to $p$ in the following way: Find a value of $t$ so that $(v+t p) \cdot p = 0$. Then the vector $(v+t p)$ will be perpendicular to $v$.

In my example, $(v+t p) = (3 + 3 t)\mathbf{i}+4 t \mathbf{j}-2t\mathbf{k}$, and $(v+t p) \cdot p = 9 + 29 t$. By choosing $t=-\frac{9}{29}$, the vector $v+t p$ is now perpendicular to $p$.

Answer 3

A suggested solution without a branch could be: Construct an array of 2 vector elements in the following way:

arr[0] = (c,c,-a-b) arr[1] = (-b-c, a,a)
int selectIndex = ((c != 0) && (-a != b)) // this is not a branch
perpendicularVector = arr[selectIndex]

If (c, c, -a-b) is zero, selectIndex is 1 and the other vector will be selected.

Answer 4

A related problem is to construct an algorithm that finds a non-zero perpendicular vector without branching. If the input vector is N = (a,b,c), then you could always choose T = (c,c,-a-b) but T will be zero if N=(-1,1,0). You could always check to see if T is zero, and then choose T = (-b-c,a,a) if it is, but this requires a test and branch. I can't see how to do this without the test and branch.

Answer 5

The dot product of two perpendicular vectors are always 0 so if you (ai+bj+ck)*(di+ej+fk)=0 you can solve for the different variables. If you have one vector than the infinite amount of perpendicular vectors will form a plane that is perpendicular to the original vector. If you know one or two of the coordinates of the desired perpendicular line than you can find the corresponding vector(s) on that plane.

Answer 6

Remember: There exist infinite vector in 3 dimension that are perpendicular to a fixed one. Now, Let v ≠0 be the vector whose is xi+yj+zk. So , v is perpendicular to the vector 3i+4j-2k. Therefore, v.(3i+4j−2k)=0. (xi+yj+zk). (3i+4j−2k)=0 3x+4y-2z=0 ……………….(1) where i.i =j.j=k.k=1.

Now, there are three unknown variable such x, y and z in (1). You can choose any two variable whatever you like. Let y=2 and z=1,
then x=-2 from (1),

One of the vector is (-2i+2j+k). Similarly, you can choose one of two variables from (1) , then find the third variable. So, you can find infinite perpendicular vectors to the vector 3i+4j-2k.