Finding a unit vector perpendicular to another vector

Question

For example we have the vector $8i + 4j - 6k$, how can we find a unit vector perpendicular to this vector?

Answer 1

Let $\vec{v}=x\vec{i}+y\vec{j}+z\vec{k}$, a perpendicular vector to yours. Their inner product (the dot product - $\vec{u}.\vec{v}$ ) should be equal to 0, therefore: $$8x+4y-6z=0 \tag{1}$$ Choose for example x,y and find z from equation 1. In order to make its length equal to 1, calculate $\|\vec{v}\|=\sqrt{x^2+y^2+z^2}$ and divide $\vec{v}$ with it. Your unit vector would be: $$\vec{u}=\frac{\vec{v}}{\|\vec{v}\|}$$

Answer 2

Congrats on 10'000+ views! I'd like to combine the above fine answers into an algorithm.

Given a vector $\vec x$ not identically zero, one way to find $\vec y$ such that $\vec x^T \vec y = 0$ is:

1. start with $\vec y' = \vec 0$ (all zeros);
2. find $m$ such that $x_m \neq 0$, and pick any other index $n \neq m$;
3. set $y'_n = x_m$ and $y'_m = -x_n$, setting potentially two elements of $\vec y'$ non-zero (maybe one if $x_n=0$, doesn't matter);
4. and finally normalize your vector to unit length: $\vec y = \frac{\vec y'}{\|\vec y'\|}.$

(I'm referring to the $n$th element of a vector $\vec v$ as $v_n$.)

Answer 3

Every answer here gives the equation $8a+4b-6c=0$. None mentions that this equation represents a plane perpendicular to the given vector. I am sure that the omission was an oversight of each respondent. But it deserves mention and emphasis. In the plane perpendicular to any vector, the set of vectors of unit length forms a circle. So answers will vary. The vectors $(-1,2,0)^t$ and $(2,0,3)^t$ can be chosen to be a basis for the solution space of the plane: solve for a, divide by 8, and let $2b$ and $3c$ be independent variables. You can divide each by its length $\sqrt{5}$ and $\sqrt{13}$ respectively, and take a trigonometric combination of them to get a general solution.

Answer 4

An automated procedure:

Take a standard vector $\vec e_k$ which is not parallel to $\vec v$, and form the cross product $\vec e_k\times\vec v$, which is guaranteed to be orthogonal to $\vec v$.

The crux of the method is to take the $\vec e_k$ which is "the less parallel" to $\vec v$, i.e. that minimizes the dot product: take the index $k$ corresponding to the smallest $|v_k|$, and you are on the safe side.

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Update:

In $d$ dimensions, take the standard vector $\vec e_k$ that forms the smallest dot product (in absolute value) with $\vec v$ and normalize the vector

$$\vec e_k-\frac{\vec e_k\vec v}{\|v\|^2}\vec v=\vec e_k-\frac{v_k}{\|v\|^2}\vec v.$$

Answer 5

I want to add an answer for only two dimensions since it's much more simple:

If our vector is $$v = ai + bj$$ We can easily calculate the normal of $v$, because the normal of a vector is perpendicular to it. $$n = -bi + aj$$ $n$ is now a perpendicular vector to the vector v, it just has to be normalized, which is explained in other answers already. $$n \rightarrow \frac{n}{\lVert n \rVert}$$

Answer 6

Two steps:

First, find a vector $a\,{\bf i}+b\,{\bf j}+c\,{\bf k}$ that is perpendicular to $8\,{\bf i}+4\,{\bf j}-6\,{\bf k}$. (Set the dot product of the two equal to 0 and solve. You can actually set $a$ and $b$ equal to 1 here, and solve for $c$.)

Then divide that vector by its length to make it a unit vector. This unit vector will still be perpendicular to $8\,{\bf i}+4\,{\bf j}-6\,{\bf k}$ .

Answer 7

A vector $v=ai+bj+ck$ is perpendicular to $w=8i+4j-6k$ if and only if $$v\cdot w=8a+4b-6c=0.$$ So for example, we could choose $a=1,b=1,c=2$, so that $v=i+j+2k$. But this is not a unit vector: $$\|v\|=\sqrt{a^2+b^2+c^2}=\sqrt{1^2+1^2+2^2}=\sqrt{6}.$$ However, for any number $t$, it is the case that $\|tv\|=|t|\cdot\|v\|$, and $(tv)\cdot w=t(v\cdot w)$. This shows us how to modify our vector $v$ to get a unit vector that still retains the property of being perpendicular to $w$. Specifically, $$u=\frac{1}{\sqrt{6}}\cdot v=\left(\frac{1}{\sqrt{6}}\right)i+\left(\frac{1}{\sqrt{6}}\right)j+\left(\frac{2}{\sqrt{6}}\right)k$$ satisfies $$\|u\|=\frac{1}{\sqrt{6}}\|v\|=\frac{1}{\sqrt{6}}\cdot\sqrt{6}=1,$$ so that $u$ is unit vector, and $$u\cdot w=\frac{1}{\sqrt{6}}(v\cdot w)=\frac{1}{\sqrt{6}}\cdot0=0,$$ so that $u$ is perpendicular to $w=8i+4j-6k$.

Answer 8

You are just looking to a vector $(x,y,z)$ s.t. $8x+4y-6z=0$. Take $(1,-2,0)$ for example, and then divide it by its norm to make it unit. Another method is that from any non-colinear vector to the first, you can apply Gramm-Schmidt process to get an orthogonal vector from the second.

Answer 9

THEORY:- Suppose 2 vectors A(given) and B,we need to find vector B such that it is Perpendicular to vector A. We also know that A.B=0 (since angle b/w them is 90° and cos(90°)=0. The below described algorithm can be applied for 2D as well as 3D vectors.

Proof:- Given A (8i+4j-6k) Total 3 steps.

Step 1:- Suppose a vector B (i+j+k).

Step 2:- Now firstly compare and divide the coefficient of each unit vector in B with that of A. As in this case, => B=(1/8)i+(1/4)j+(1/-6)k.

Step 3:- Now multiply any one of the three coefficient of unit vector in B with (-2). As in this case i have multiplied (-2) with the coefficient of j.It can be done with i or k also. ** => B=(1/8)i+(1/4)j+(1/3)k. Now for the Unit Vector we can divide vector B with its magnitude, which would be in this case, => |B|= √(1/8)² +(1/4)² +(1/3)²

**=> Unit (B)=B/|B|

** (B) IS THE VECTOR PERPENDICULAR TO A. ** Unit(B) IS THE REQUIRED UNIT VECTOR. Verification:-This can be verified as A.B=0

Proved. :)

Answer 10

In addition to Yves Daoust's answer,

In d dimensions, take any random vector $w$ which is not parallel to $v$, then $$u=w-\frac{v^Tw}{||v||^2}v$$ is orthogonal to the vector $v$. It doesn't have to be a standard vector (or that forms the smallest dot product).

Answer 11

You need to find a*i + b*j + c* k so that the dot product of it with 8i +4j -6k is 0.

It means 8a + 4b - 6 c = 0.

You need to choose a,b,c satisfy above. For example, you can choose a = 1, b = 1, c = 2.