# Write the dot product of $(1, 4, 5)$ and $(x, y, z)$as a matrix multiplication

I'm very confused by this question in practice:

Write the dot product of $$(1, 4, 5)$$ and $$(x, y, z)$$ as a matrix multiplication $$Ax$$. (The matrix A should only have one row). The solutions to $$Ax = 0$$ lie on a ___ perpendicular to the vector ___. The columns of $$A$$ are vectors in only __-dimensional space.

What I currently think is the dot product as a matrix multiplication $$Ax$$ is $$Ax = \begin{bmatrix}1&4&5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}$$ where $$A$$ is $$\begin{bmatrix}1&4&5\end{bmatrix}$$. Hence the solutions to $$Ax=0$$ is the same as $$x+4y+5z=0$$. But what do those solutions lie on ___ perpendicular to what vector? Are the columns of A vectors in only 1-dimensional space?

Answer:

Thanks to @pipe's answer. The solutions to $$x+4y+5z=0$$ lie on a plane (because it's a three-variable equation) perpendicular to $$\begin{bmatrix}1&4&5\end{bmatrix}$$. Because for any solution $$(x, y, z)$$, its dot product with $$\begin{bmatrix}1&4&5\end{bmatrix}$$ is always zero. According to the formula, $$cos\theta=\frac{\vec{a}\vec{b}}{||\vec{a}||||\vec{b}||}$$, the numerator would be $$0$$, which makes $$cos\theta=0$$, therefore $$\theta=90^\circ$$. The columns of $$A$$ are vectors in only 1-dimensional space before there's only one entry in every column of $$A$$.

• You've written the dot product as a matrix product correctly. The column vectors of length 3 form a three dimensional space. The ones you are interested in form the two dimensional subspace perpendicular to $[1,4,5]$. Sep 28 at 14:06
• In your question, $x$ is both a $3\times 1$ matrix and a number in such matrix? Sep 28 at 14:06
• @peterwhy I think so. I'm so confused Sep 28 at 14:09

## 1 Answer

The solutions to $$Ax=0$$ lie on a plane perpendicular to the vector $$(1,4,5)$$. The columns of $$A$$ are vectors in $$1$$-dimensional space.

Indeed, $$ax+by+cz=d$$ is a plane perpendicular to $$(a,b,c)$$.

• oh so is this "𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑑 is a plane perpendicular to (𝑎,𝑏,𝑐)." something that is always true? Sep 28 at 14:13
• Yes. It's easy to see that if you take two solutions, $(x_0,y_0,z_0),(x_1,y_1,z_1)$, their difference dots $(a,b,c)$ to zero.
– ACME
Sep 28 at 14:19
• it's true for whatever d is? Sep 28 at 14:29
• Yes. Changing $d$ gives a different plane with normal $(a,b,c)$.
– ACME
Sep 28 at 14:37
• Yes, I reckon if $A$ is $n×1$ then $Ax=0$ will be a hyperplane in $n$-space. The equation reduces the dimension by $1$.
– ACME
Sep 28 at 19:27