How can I determine the vector the the solutions to $A\overrightarrow { x } = 0$ lie on?

This question was taken from MIT OCW and Introduction to Linear Algebra by Gilbert Strang

Write the dot product of $(1,4,5)$ and $(x,y,z)$ as a matrix multiplication of $A\overrightarrow { x }$. The matrix $A$ has one row. The solutions to $A\overrightarrow { x } = 0$ lie on a ____ perpendicular to the vector ____. The columns of $A$ are only in ____-dimensional space.

What I understand thus far and what I need clarification on:

1) The solutions to $A\overrightarrow { x } = 0$ lie on a plane. However, the answer saids: "...on a plane in three dimensions". Why?

2) ...perpendicular to the vector ____. I don't understand this part. The solution to this is the $0$ vector, no?

3) The columns of $A$ are only in one-dimensional space, since matrix $A$ is $[1,4,5]$

The dot product of n-tuples (scroll down to Tuples under Vectors) is given by $$\vec a \cdot \vec b = (a_1, a_2, \dots, a_n) \cdot (b_1, b_2, \dots, b_n) = a_1b_1 + a_2b_2 + \cdots + a_nb_n$$

This can be calculated by converting the vectors $\vec a$ and $\vec b$ into column matrices: $\mathbf a = \pmatrix{a_1 \\ a_2 \\ \vdots \\ a_n}$ and $\mathbf b = \pmatrix{b_1 \\ b_2 \\ \vdots \\ b_n}$. Then $$\mathbf a^T\mathbf b = \pmatrix{a_1b_1 + a_2b_2 + \cdots + a_nb_n} = \pmatrix{\vec a \cdot \vec b}$$

If we think of $n$-tuples as arrow vectors in Euclidean space, then we can test for perpendicularity (denoted $\vec a\ \bot\ \vec b$) of two vectors by using the dot product and noting that $$\vec a\ \bot\ \vec b \iff \vec a \cdot \vec b = 0$$

Now let's look at the dot product of $(1,4,5)$ and $(x,y,z)$. Your question asks about solutions to $(1,4,5)\cdot (x,y,z)=0$. Using the above relationship between the dot product and perpendicularity we can see that the solution set to this equation is all of the vectors perpendicular to $(1,4,5)$.

I don't know how far you've gotten into Linear Algebra, but because $3$-tuples are members of $\Bbb R^3$, which is a $3$-dimensional space, the subspace of $\Bbb R^3$ perpendicular to a given nonzero vector will be a $2$-dimensional space: i.e. a plane. That's what the solution set is here.

The dot product satisfies $\boldsymbol{x} \cdot \boldsymbol{y} = \|\boldsymbol{x}\|\|\boldsymbol{y}\|\cos \theta$, where $\|\boldsymbol{x}\|$ denotes the Euclidean length of the vector $\boldsymbol{x}$ and $\theta$ is the angle between them. Thus, if $\boldsymbol{x} \cdot \boldsymbol{y}= 0$ then the two vectors are perpendicular. The condition that $A \cdot \boldsymbol{x} = 0$ thus means that $A = (1,4,5)$ is perpendicular to $\boldsymbol{x}$.
• So, the solutions to $A\overrightarrow { x } = 0$ are perpendicular to $\overrightarrow { x }$? Nov 21, 2015 at 14:07
• The solutions are the possible values of $\vec x$. The solutions are perpendicular to $(1,4,5)$.