All vectors in $\mathbb R^4$ orthogonal to two vectors

Question

Problem statement: Find all vectors in $\mathbb R^4$ that are orthogonal to the two vectors $u_1=(1,2,1,3)$ and $u_2=(2,5,1,4)$.

Progress:

a) Denote a vector $u_3=(v_1,v_2,v_3,v_4)$ My desire is to determine $u_3$ so that $\left \langle u_1,u_3 \right \rangle=\left \langle u_2,u_3 \right \rangle=0$

$\left \langle u_1,u_3 \right \rangle=(1,2,1,3)*(v_1,v_2,v_3,v_4)=v_1+2v_2+v_3+3v_4=0$

$\left \langle u_2,u_3 \right \rangle=(2,5,1,4)*(v_1,v_2,v_3,v_4)=2v_1+5v_2+v_3+4v_4=0$

Thus I end up with a system of equations I could not solve.

b) I know that the cross product of two vectors $a$ and $b$ results in a vector orthogonal to $a$ and $b$ that cannot be applied in $\mathbb R^4$.

Answer 1

For a), you have the correct system of equations. Do you know the Gauss-Jordan elimination algorithm? When you apply it to your system of equations, you will find that there are two free variables, so you will have a two-dimensional subspace.

For b), you are correct that the cross-product is only defined for vectors in $\mathbb{R}^3$.

Intuitively, to be orthogonal to the given vectors $u_1$ and $u_2$ is equivalent to being orthogonal to the plane spanned by $u_1$ and $u_2$. In $\mathbb{R}^4$, the set of all vectors orthogonal to a plane has dimension $4 - 2 = 2$, so it is also a plane.

SPOILER: To check your work, the canonical basis obtained by solving your system of equations is $w_1 = (-3,1,1,0)$ and $w_2 = (-7,2,0,1)$. I'd be happy to give more details if you're stuck, but it sounds like you are close to solving it on your own.

Answer 2

In the given system, you have a matrix $ A= \begin{bmatrix} 1 & 2 & 1 & 3 \\ 2 & 5 & 1 & 4 \end{bmatrix} $ and you need to find other two vectors in $\mathbb{R}^4$, so you can find the Null Space of $A$, as the Null Space $N(A)$ is orthogonal to Row Space $C(A^T)$, so all the vectors in the Null Space will be orthogonal to every vector in Row Space.

Step 1: Find $R = rref(A)$. \begin{equation} R = rref(A) = \begin{bmatrix} 1 & 0 & 3 & 7 \\ 0 & 1 & - 1 & -2 \end{bmatrix} \end{equation} Step 2: Find $Rx=0$ where $x = \begin{bmatrix} x_1 & x_2 & x_3 & x_4 \end{bmatrix}^T$

Here $x_3$ and $x_4$ are free variables, so we need to solve the system \begin{equation} \begin{bmatrix} 1 & 0 & 3 & 7 \\ 0 & 1 & - 1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \end{equation}

a) Putting $x_3=1$ and $x_4=0$, we get $x_1=3$ and $x_2=1$, so the first vector $x = \begin{bmatrix} -3 & 1 & 1 & 0 \end{bmatrix}$ b) Putting $x_3=0$ and $x_4=1$, we get $x_1=-7$ and $x_2=2$, so the second vector $x = \begin{bmatrix} -7 & 2 & 0 & 1 \end{bmatrix}$

Thus the complete system will be \begin{equation} B = \begin{bmatrix} 1 & 0 & 3 & 7 \\ 0 & 1 & - 1 & -2 \\ -3 & 1 & 1 & 0 \\ -7 & 2 & 0 & 1 \end{bmatrix} \end{equation}

Here both the third and fourth rows are perpendicular to both first and second rows, this can be verified by taking dot products.