# Finding a unit vector perpendicular to another vector

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

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

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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.

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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 $w=8i+4j-6k$.

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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 $8\,{\bf i}+4\,{\bf j}-6\,{\bf k}$ .

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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: $\frac{\vec y}{\|\vec y\|}.$

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

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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.

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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.

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That does not give a unit vector. –  Zev Chonoles Apr 17 '12 at 22:34