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.
$\endgroup$ 1118 Answers
$\begingroup$There exists an infinite number of vectors 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: $$v=((4a+2b)\mathbf{i}-3a\mathbf{j}+3b\mathbf{k}) \hspace{10 mm} a,b \in \mathbb{R}$$
$\endgroup$ 10 $\begingroup$Take cross product with any vector. You will get one such vector.
$\endgroup$ 3 $\begingroup$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.
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 $(x+t p) \cdot p = 0$. Then the vector $v=x+t p$ will be perpendicular to $p$.
In my example, $(x+t p) = (3 + 3 t)\mathbf{i}+4 t \mathbf{j}-2t\mathbf{k}$, and $(x+t p) \cdot p = 9 + 29 t$. By choosing $t=-\frac{9}{29}$, the vector $v=x+t p$ is now perpendicular to $p$.
$\endgroup$ 5 $\begingroup$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.
For any nonzero vector $(a,b,c)$, the three of $(0,c,-b),(-c,0,a)$ and $(-b,a,0)$ are orthogonal to it.
To avoid the "parallel case", you can choose the one with the largest squared modulus, among $c^2+b^2, c^2+a^2$ and $b^2+a^2$, or the one with the two largest absolute components or simply one with the largest absolute component. Choosing the largest will also optimize numerical stability.
In the given case, $(-4,3,0)$.
Update:
The largest squared modulus also corresponds to the smallest (absolute) component.
$\endgroup$ $\begingroup$Short answer: the vector $(s_z\,(z + s_z) - x^2, -x y, -x\,(z + s_z))$ with $s_z := \text{sign}(z) \, \|(x,y,z)\|$ is orthogonal to the vector $(x,y,z)$.
Note that we assume that $\text{sign}(x)$ is defined as $+1$ for $x \ge 0$ and as $-1$ otherwise.
Let $(x,y,z)$ be a vector with norm s and z > -s then the following matrix is an orthogonal basis where every basis vector has norm s:
$\left( \begin{array}{ccc} s - \frac{x^2}{z+s} & -\frac{x y}{z+s} & x \\ -\frac{x y}{z+s} & s - \frac{y^2}{z+s} & y \\ -x & -y & z \\ \end{array} \right)$
There are two notable cases if z = -s:
- The vector is of form $(0,0,z)$ with z < 0 and we can simply invert it before applying the formula above. As shown below this can be exploited to get a branch-free implementation.
- The vector is the zero vector $(0,0,0)$. "perpendicular" doesn't make much sense in case of the null vector. If you interpret it as "dot product is zero" than you can just return the zero vector.
We can deal with these two problems as follows:
Let's look at the first vector: $(s - \frac{x^2}{z+s}, -\frac{x y}{z+s}, -x)$. The singularity at $(0,0,-1)$ can be avoided by inverting the input vector and then inverting the result which gives: $(-s - \frac{x^2}{z-s}, -\frac{x y}{z-s}, -x)$.
Following this idea we can set $s_z := \text{sign}(z) \, s$ and compute an orthogonal basis vector for any non-null vector $(x,y,z)$ as:
$(s_z - \frac{x^2}{z + s_z}, -\frac{x y}{z + s_z}, -x)$
This leads to a nice branch-free C++ implementation for a normalized vector:
Vector3 OrthoNormalVector(double x, double y, double z) { const double g = std::copysign(1., z); const double h = z + g; return Vector3(g - x*x/h, -x*y/h, -x);
}Check the implementation of copysign on your platform to make sure that copysign(1., 0.) returns 1 and not 0.
For an arbitrary vector, not necessarily normalized, we can use a little trick to get an orthogonal vector: we scale the vector by the factor $z+s_z$ to get:
$(s_z\,(z + s_z) - x^2, -x y, -x\,(z + s_z))$
This vector is still orthogonal to the original vector $(x,y,z)$ as it was just scaled by a factor. It also has zero norm if and only if the norm of the original vector is 0.
This leads again to a branch-free implementation:
Vector3 OrthogonalVector(double x, double y, double z) { const double s = std::sqrt(x*x + y*y + z*z); const double g = std::copysign(s, z); // note s instead of 1 const double h = z + g; return Vector3(g*h - x*x, -x*y, -x*h);
}One way to do this is to express the vector in terms of a spherical coordinate system. For example
$$ \boldsymbol{e}= \pmatrix{a \\ b \\ c} = r \pmatrix{ \cos\varphi \cos\psi \\ \sin\varphi \cos\psi \\ \sin\psi} $$
where $r=\sqrt{a^2+b^2+c^2}$, $\tan(\varphi) = \frac{b}{a}$ and $\tan{\psi} = \frac{c}{\sqrt{a^2+b^2}}.$
Provided that $a \neq 0$ or $b \neq 0$ then
A choice of two orthogonal vectors can be found with $$ \begin{aligned} \boldsymbol{n}_1 & = \frac{{\rm d} \boldsymbol{e}}{{\rm d} \varphi} = r\pmatrix{-\sin \varphi \cos\psi \\ \cos\varphi \cos\psi \\ 0}& \boldsymbol{n}_2 & = \frac{{\rm d} \boldsymbol{e}}{{\rm d} \psi} = r\pmatrix{-\cos\varphi \sin\psi \\ -\sin\varphi \sin\psi \\ \cos\psi} \end{aligned}$$
Of course, any non-zero linear combination of these two vectors is also orthogonal
$$ \boldsymbol{n} = \cos(t) \boldsymbol{n}_1 + \sin(t) \boldsymbol{n}_2 $$
where $t$ is an rotation angle about the vector $\boldsymbol{e}$.
Put it all together to make a family of orthogonal vectors in terms of $t$ as
$$ \boldsymbol{n} = \pmatrix{-b \cos(t) - \frac{a c}{\sqrt{a^2+b^2}} \sin(t) \\ a \cos(t) - \frac{b c}{\sqrt{a^2+b^2}} \sin(t) \\ \sqrt{a^2+b^2} \sin(t)} $$
For $\boldsymbol{e} = \pmatrix{3 & 4 & -2}$ the aboves gives
$$ \boldsymbol{n} = \pmatrix{ \frac{6}{4} \sin(t)-4 \cos(t) \\ 3 \cos(t) - \frac{8}{5} \sin(t) \\ 5 \sin(t) } \longrightarrow \begin{cases} \boldsymbol{n} = \pmatrix{-4 & 3 & 0} & t =0 \\ \boldsymbol{n} =\pmatrix{\frac{6}{5} & \frac{8}{5} & 5} & t = \frac{\pi}{2} \end{cases} $$
For the case when $a=0$ and $b=0$ then you know can assign the perpendicular somewhat arbitrarily with
$$ \boldsymbol{n}_1 = \pmatrix{1 \\ 0 \\ 0} $$
and
$$ \boldsymbol{n}_2 = \pmatrix{0 \\ 1 \\ 0} $$
for the general solution
$$\boldsymbol{n} = \cos(t) \boldsymbol{n}_1 + \sin(t) \boldsymbol{n}_2$$
$\endgroup$ 7 $\begingroup$This branch-free algorithm is $\operatorname{sqrt}$-free and trig-free:
$$ \begin{aligned} \begin{bmatrix} \operatorname{copysign}\left(z,x\right) \\ \operatorname{copysign}\left(z,y\right) \\ -\operatorname{copysign}\left(|x|+|y|,z\right) \\ \end{bmatrix} \end{aligned} $$
An equivalent form
This alternative avoids the 2 $\operatorname{abs}$at the cost of an additional $\operatorname{copysign}$:
$$ \begin{aligned} \begin{bmatrix} \operatorname{copysign}\left(z,x\right) \\ \operatorname{copysign}\left(z,y\right) \\ -\operatorname{copysign}\left(x,z\right) -\operatorname{copysign}\left(y,z\right) \\ \end{bmatrix} \end{aligned} $$
Properties
Let $L_\text{i}$be the length of the input and $L_\text{o}$be the length of the output:
$$ \begin{aligned} L_\text{i} \le L_\text{o} \le \sqrt{2} L_\text{i} \end{aligned} $$
which holds for both forms above.
A note about the function $\operatorname{copysign}$
Many platforms offer a function$\operatorname{copysign}\left(a,b\right)$whose return value has the magnitude of $a$and the sign of $b$. Despite the following mathematical definition, its implementation can be branch-free using bitwise operations:
$$ \begin{aligned} \operatorname{copysign}\left(a,b\right) &= \begin{cases} |a|&\text{for }b\ge0 \\ -|a|&\text{for }b<0 \\ \end{cases} \end{aligned} $$
If $\operatorname{copysign}$ is not available
The $\operatorname{copysign}(a,b)$ function is preferred because it is non-vanishing. However some platforms only offer a $\operatorname{sign}(b)$function which vanishes for $b=0$:
$$ \begin{align} \operatorname{sign}(b) &= \begin{cases} 1 &\text{for } b > 0 \\ 0 &\text{for } b = 0 \\ -1 &\text{for } b < 0 \\ \end{cases} \\ \end{align} $$
Fortunately an alternative exists. The desired non-vanishing functionality can be obtained by "nudging" the output and nesting the result in another call:
$$ \begin{align} \operatorname{sign}\left[\operatorname{sign}(b)+0.5\right] &= \begin{cases} 1 &\text{for } b \ge 0 \\ -1 &\text{for } b < 0 \\ \end{cases} \\ \end{align} $$
This leads to the following form of the perpendicular to vector $(x,y,z)$for platforms with no $\operatorname{copysign}(a,b)$ function:
$$ \begin{align} \begin{bmatrix} s_{xz}z \\ s_{yz}z \\ -s_{xz}x-s_{yz}y \end{bmatrix} \\ \end{align} $$
where:
$$ \begin{align} s_{xz} &= \operatorname{sign}\left\{ \left[ \operatorname{sign}(x)+0.5 \right] \left[ \operatorname{sign}(z)+0.5 \right] \right\} \\ s_{yz} &= \operatorname{sign}\left\{ \left[ \operatorname{sign}(y)+0.5 \right] \left[ \operatorname{sign}(z)+0.5 \right] \right\} \\ \end{align} $$
$\endgroup$ $\begingroup$A geometric solution would be as follows. The plane $3x+4y-2z=0$ is perpendicular to the vector $3i+4j−2k$. Any vector in that plane is thus perpendicular this vector. Thus you may choose any $x$, $y$ and $z$ that lie in the plane $3x+4y-2z=0$ and the resulting $xi+yj+zk$ will be perpendicular to $3i+4j−2k$,
$\endgroup$ $\begingroup$Given $n-1$ linearly independent vectors, $\{v_j\}_{j=1}^{n-1}$ in $\mathbb{R}^n$, we can find a non-zero vector, $u$, perpendicular to all of them.
If we set $$ \begin{align} u_1&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&1\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&0 \end{bmatrix}\\ u_2&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&0\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&0 \end{bmatrix}\\ &\vdots\\ u_n&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&0\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&1 \end{bmatrix}\\ \end{align} $$ then $$ u\cdot w=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&w_1\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&w_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&w_n \end{bmatrix} $$ If we replace $w$ by any of the $v_j$, the determinant will be $0$ because of duplicate columns; thus, $u\cdot v_j=0$.
$\{v_j\}_{j=1}^{n-1}$ cannot span $\mathbb{R}^n$, so there must be some $v_n$ that is not in the span of $\{v_j\}_{j=1}^{n-1}$. This means that $\{v_j\}_{j=1}^n$ are independent, and so $$ \begin{align} u\cdot v_n&=\det\begin{bmatrix} v_{1,1}&v_{2,1}&\cdots&v_{n-1,1}&v_{n,1}\\ v_{1,2}&v_{2,2}&\cdots&v_{n-1,2}&v_{n,2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ v_{1,n}&v_{2,n}&\cdots&v_{n-1,n}&v_{n,n} \end{bmatrix}\\ &\ne0 \end{align} $$ In particular, $u\ne0$.
$\endgroup$ 4 $\begingroup$The vectors perpendicular to $(3,4,-2)$ form a two dimensional subspace, the plane $3x+4y-2z=0$, through the origin.
To get solutions, choose values for any two of $x,y$ and $z$, and then use the equation to solve for the third.
The space of solutions could also be described as $V^{\perp}$, where $V=\{(3t,4t,-2t):t\in\Bbb R\}$ is the line (or one dimensional vector space) spanned by $(3,4-2)$.
$\endgroup$ $\begingroup$Another way to find a vector $\vec{v}$ for a given $\vec{u}$ such that$$ \vec{u}\cdot\vec{v}=0 $$is to use an antisymmetric matrix $A$ ($A^\top=-A$) defined as follow$$ A_{ij}u_iu_j=0\qquad(\text{sum over }ij). $$In two dimension $A$ is$$ A=\begin{pmatrix} 0&1\\ -1&0\\ \end{pmatrix}. $$In three dimension $A$ is$$ A=\begin{pmatrix} 0&1&1\\ -1&0&1\\ -1&-1&0\\ \end{pmatrix}. $$In 2D only one such vector $\vec{v}=A\vec{u}$ exist, while in 3D you can apply the same matrix to the sum $\vec{u}+\vec{v}$ finding a vector perpendicular to the plane given by the other two vectors.
2D
The matrix $A$ can be calculated as follow$$ A_{ij}u_iu_j=A_{11}u_1^2+(A_{12}+A_{21})u_1u_2+A_{22}u_2^2. $$One way is to set $A_{11}=0=A_{22}$ and $A_{21}=-A_{12}$.
3D
Again$$ A_{ij}u_iu_j=A_{11}u_1^2+(A_{12}+A_{21})u_1u_2+A_{22}u_2^2+(A_{13}+A_{31})u_1u_3+(A_{23}+A_{32})u_2u_3+A_{33}u_3^2, $$and setting $A_{11}=A_{22}=A_{33}=0$ and $A_{21}=-A_{12}$, $A_{31}=-A_{13}$ and $A_{23}=-A_{32}$.
$\endgroup$ $\begingroup$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
Example:
Let
$\vec{a}$ = ( 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
... as Rebecca said: << Keep in mind there will be an infinite number of perpendicular vectors >> ... Here the pdf source
$\endgroup$ 4 $\begingroup$All vectors perpendicular to the given vector form a plane. If $v_1$ and $v_2$ are perpendicular to the given vector $v = 3i +4j -2k$, then the dot products $v\cdot v_1 =0$ and $v\cdot v_2 = 0$. If $v_1 = 2i -j + k$ and $v_2 = 2i +j +5k$, then a plane formed by any vector $v_3 = av_1 +bv_2$; where $a$ and $b$ are scalars, will be normal to the given vector $v$.
$\endgroup$ $\begingroup$Remember: There exist infinite vector in 3 dimension that are perpendicular to a fixed one. Now, Let $v\neq 0$ be the vector whose is $xi+yj+zk$. So , $v$ is perpendicular to the vector $3i+4j-2k$. Therefore, $v\cdot\langle 3i+4j−2k\rangle=0$. $$ \langle xi+yj+zk\rangle\cdot \langle3i+4j−2k\rangle =0 $$ so $3x+4y-2z=0$ (1) where $i\cdot i =j\cdot j=k\cdot k=1$.
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 $3i+4j-2k$.
$\endgroup$ $\begingroup$A vector perpendicular to the given vector A can be rotated about this line to find all positions of the vector. To find them,
if $ A \cdot B =0 $ and $ A \cdot C =0 $ then $ B,C $ lie in a plane perpendicular A and also $ A \times ( B \times C ) $= 0, for any two vectors perpendicular to A. (Last equation typo edited late)
$\endgroup$ $\begingroup$The dot product of two perpendicular vectors are always $0$ so if you$(ai+bj+ck)\cdot(di+ej+fk)=0$ 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.
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