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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 $A$.

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  • $\begingroup$ 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 $[1,4,5]$. $\endgroup$ Sep 28 at 14:06
  • $\begingroup$ In your question, $x$ is both a $3\times 1$ matrix and a number in such matrix? $\endgroup$
    – peterwhy
    Sep 28 at 14:06
  • $\begingroup$ @peterwhy I think so. I'm so confused $\endgroup$
    – d0nut
    Sep 28 at 14:09

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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 $(a,b,c)$.

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  • $\begingroup$ oh so is this "๐‘Ž๐‘ฅ+๐‘๐‘ฆ+๐‘๐‘ง=๐‘‘ is a plane perpendicular to (๐‘Ž,๐‘,๐‘)." something that is always true? $\endgroup$
    – d0nut
    Sep 28 at 14:13
  • $\begingroup$ 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. $\endgroup$
    – ACME
    Sep 28 at 14:19
  • $\begingroup$ it's true for whatever d is? $\endgroup$
    – d0nut
    Sep 28 at 14:29
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    $\begingroup$ Yes. Changing $d$ gives a different plane with normal $(a,b,c)$. $\endgroup$
    – ACME
    Sep 28 at 14:37
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    $\begingroup$ Yes, I reckon if $A$ is $n×1$ then $Ax=0$ will be a hyperplane in $n$-space. The equation reduces the dimension by $1$. $\endgroup$
    – ACME
    Sep 28 at 19:27

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