I've got a problem with two vectors A=(1,2,-3) and B=(2,-1,3), where I have to find the coordinates for a vector that is perpendicular to both A and B. I know I can use the cross product method for doing this, but without using this, how would one go about finding the solution?

I've tried using the idea that $A\cdot N = 0$ if $N$ is perpendicular to $A$ as a means for doing this, but I get the following:

$$A\cdot N = 0$$ $$B\cdot N = 0$$

...and as far as I can tell, I cannot do anything else. These are just the required final properties for the vectors to be orthogonal.

My approach is to use some variables N=(x,y,z), to get:

$$A\cdot N = x+2y-3z = 0$$ $$B\cdot N = 2x-y+3z = 0$$

...and then know that the line where the resulting planes intersect is the perpendicular vector, so this means these two equations must be in balance:

$$x+2y-3z = 2x-y+3z$$


$$-x+3y-6z = 0$$

...however, this just produces another plane that intersects along the same line, which is the same for further subtracting or adding any of these subtracted or added to each other, so for example, this holds true with the following plane as well:

$$3x+y = 0$$

From here, I could plug and chug to eliminate variables and then find a value for x, y, and z that satisfy these equations, but I am curious if there is another less brutal approach for this, without resorting to cross product (or is the cross product indeed the only alternative approach)?

This question stems from me observing the finesse of properties for showing orthogonal and parallel vectors, which just involve taking dot products and/or subtracting a vector located at a point and finding a projection along it using ratio of dot products, etc. These approaches are very clean, and thats what I'm looking for here, if it exists.

Even if there is a no basic formula (besides cross product) or simple method for finding this, I'd be satisfied with any approach that simplifies the use of elimination and plugging and chugging.


The equations you have, which are basically $$ \begin{pmatrix} A^T \\ B^T \end{pmatrix} N = 0, $$ really are the most general conditions for a perpendicular vector (and in more than three dimensions, this is basically all you can say). So you're looking for the kernel of the matrix $\begin{pmatrix} A^T \\ B^T \end{pmatrix}$. Since the kernel is a subspace, any vector will do.

The "canonical" way to find such a vector is probably to use something like Gram-Schmitt: take a vector $N_0$ not in the span of $\{A,B\}$ (i.e. linearly independent). Also, wlog $A \cdot B=0$, else take $$ B' = B-\frac{B \cdot A}{\lVert A \rVert^2}A: $$ note that $B' \cdot A = B \cdot A - (B \cdot A)(A \cdot A)/\lVert A \rVert^2 =0$.

Then if we take $$ 0 \neq N' := N_0-\frac{N_0 \cdot A}{\lVert A \rVert^2}A-\frac{N_0 \cdot B'}{\lVert B' \rVert^2}B', $$ we have $$ N' \cdot A = N_0 \cdot A -\frac{N_0 \cdot A}{\lVert A \rVert^2}A \cdot A -0=0 $$ and $$ N' \cdot B' = N_0 \cdot B' -0 -\frac{N_0 \cdot B'}{\lVert B' \rVert^2}B' =0. $$

On the other hand, you may wish to look up the wedge product, the Hodge dual and bivectors.

  • $\begingroup$ I think you wanted $N_0$ to be outside the span of $A$ and $B$, right? In general, $(A+B)\cdot A \neq \|A+B\| \|A\|$, but $A+B$ is not what you want. $\endgroup$
    – David K
    Apr 12 '15 at 13:38
  • $\begingroup$ @DavidK Yes, bit of unjustified streamlining there... I'll fix it. $\endgroup$
    – Chappers
    Apr 12 '15 at 14:32
  • $\begingroup$ Gram-Schimtt orthogonalization was the answer I expected to see. $\endgroup$ May 3 '20 at 0:45

I don't know what you are asking for. If you wish to find a perpendicular vector there are many ways to do that. If it is not a vector you want but a magnitude there are faster ways to go about that.

make a matrix with your vectors then multiply   
    1 2 -3  * x          
    2 -1 3    y     = 0  leading to an augmented matrix  1 2 -3|0   
              z                                          2 -1 3|0  

Use Gauss: R2-2*R1 ; R1+R2(2/5) ;   R2/-5    
1 2 -3  |0                
0 -5 -3 |0             

1 0 -6/5 |0  
0 1 3/5  |0   

solve for variables x=6/5z y=-3/5z were z can be any real number.Any vector in this plane is normal.

for the magnitudes of cross products think about finding the area of a parallelogram. If you want to use dot product sqrt(|a|^2*|b|^2-(ab)^2) you can get this with Gram, Lagrange, or with |a||b|sin(theta) Pythagorean


if you row reduce $$\pmatrix{1&2&-3\\2&-1&3} \to \pmatrix{1&0&0.6\\0&1&-1.8}$$ we have a free variable $z.$

Setting $z = 5,$ we get $y = 9, x = -3.$ Therefore any multiple of $$\pmatrix{-3\\9\\5} \perp \text{span}\left\{\pmatrix{1\\2\\-3}, \pmatrix{2\\-1\\3}\right\}.$$


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