# “best practice” for finding perpendicular vectors

I'm working through some examples in Strang (Ed. 3) and the solution manual.

A couple questions in a row now have been "find vectors perpendicular to the given vector as well as perpendicular to each other"

The 3d one I solved by drawing out a 3-d matrix. The original vector was (1,0,1). I visually draw one perpendicular to that (-1,0,1). And then visually inspect to see the next perpendicular vector I can see (0,1,0)

Then I was given a 4d example -- find three vectors perpendicular to (1,1,1,1) and each other.

In this case I have a hard time drawing it. What's the best way to approach a solution for this kind of problem (nothing stopping the text from asking me for a 6-dimension example next)? Should I just find an arbitrary perpendicular vector. And then for the second vector work out multiple equations for (A.v2) and (v1.v2) and solve for the v2 parameters? and then keep doing that till I can generate a v3? Or should I be smarter in my intuition here.

Feeling dumb. I don't think I have a foolproof approach to this problem class.

• – mvw Oct 10 '14 at 11:36

Use the dot product, not visual perpendicularity.

In particular, $(1,-1,0,0)\cdot(1,1,1,1) = 0$, so these two vectors must be perpendicular. $(0,0,1,-1)$ also works for the same reason. Then, take $(1,1,-1,-1)$.

If you don't want to guess, you can take the vectors $(1,1,1,1),(1,0,0,0),(0,1,0,0),(0,0,1,0)$ and apply the Gram-Schmidt process.

Another way to think about this: we start by looking for a vector $v_2$ in the kernel of $v_1$. We then look for a vector $v_3$ in the kernel of $$\pmatrix{v_1^T\\v_2^T}$$ We then look for a vector $v_4$ in the kernel of $$\pmatrix{v_1^T\\v_2^T\\v_3^T}$$ And so on.

In this case, I haven't read about the Graham-Schmidt process yet, so I wound up just using multiple equations for their constraints.

so if A is (1,1,1,1),

first find vector X. The only constraint for that is x1+x2+x3+x4 = 0. so I just pick (-1, 1, 0, 0)

The second vector Y has more constraints. a) (1)y1+(1)y2+(1)y3+(1)y4 = 0 x) (-1)(y1) + (1)(y2) + 0 + 0 = 0: y1=y2

Y = (0, 0, 1, -1) (error towards the side of choosing more zeros in a vector)

The third vector Z has the most constraints.

a) z1+z2+z3+z4 = 0 x) z1(-1) + z2(1) + 0 + 0 = 0: z2 = z1 y) z1(0) + z2(0) + z3(1) + z4(-1) = 0: z3=z4

Z = (-1, -1, 1, 1)