Question

Consider the vector v = (0;3;4) in 3D. Find unit vectors u1 and u2 which are perpendicular to v in 3D.

Answer 1

Do you know the cross product?

Answer 2

Notice that \(\mathbf{u}_1=(1,0,0)\) is perpendicular. That one is kinda a freebie.
Then you wanna do the cross product: \[
\mathbf{v}\times \mathbf{u}_1 = (0,3,4)\times (1,0,0)=\mathbf{w}
\]And you make \(\mathbf{w}\) a unit vector: \[
\mathbf{u}_2=\frac{\mathbf{w}}{\|\mathbf{w}\|}
\]

Answer 3

The cross product of two vectors gives you a perpendicular vector.

Answer 4

I know the cross product, but why u1=(1,0,0) is perpendicular?

Answer 5

Well you can tell that if you do the dot product it is \(0\).

Answer 6

\[
\mathbf{v}\cdot \mathbf{u}_1 = (0,3,4)\cdot (1,0,0) = 0\cdot 1+3\cdot 0+4\cdot 0 =0+0+0=0
\]

Answer 7

Basically if you have some vector like... \[
\mathbf{a} = (a_1, a_2,\dots,a_k,\dots,a_n)
\]And you have \(a_k=0\). It will be perpendicular to some vector \[
\mathbf{b} = (b_1, b_2,\dots,b_k,\dots,b_n)
\]Where \(b_k=1\) and all other components are \(0\): \(\forall i\ne k:\;b_i=0\)

Answer 8

Now I am confused as the question says there are u1 and u2.

Answer 9

Or do we have infinite solutions?

Answer 10

There are infinite solutions.

Answer 11

\(\mathbf{v}\) is unchanged and yet we can rotate our unit vectors an infinitesimal amount.

Answer 12

Got it. Thank you very much wio!

Answer 13

Then how to present the answer saying there are infinite solutions in 3D which are
perpendicular to v?

Answer 14

up

Answer 15

can somebody help me please?