### Instruction: There exist vectors $\mathbf{a}$ and $\mathbf{b}$ such that
\[\mathbf{a} + \mathbf{b} = \begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix},\]where $\mathbf{a}$ is parallel to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$ and $\mathbf{b}$ is orthogonal to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.$  Find $\mathbf{b}.$.### Response: This problem involves some vector algebra, so I'll start by reviewing some basic facts and definitions.
Two vectors are parallel if they are scalar multiples of each other, that is, $\mathbf{a} = k \mathbf{c}$ for some constant $k.$
Two vectors are orthogonal if their dot product is zero, that is, $\mathbf{b} \cdot \mathbf{c} = 0.$
The dot product of two vectors is defined as $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3,$ where $a_1, a_2, a_3$ are the components of $\mathbf{a}$ and $b_1, b_2, b_3$ are the components of $\mathbf{b}.$
The dot product has some useful properties, such as being commutative, distributive, and linear.
One property that I think will be helpful for this problem is that the dot product of a vector with itself is equal to the square of its magnitude, that is, $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2.$
The magnitude of a vector is defined as the square root of the sum of the squares of its components, that is, $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.$
Now that I have reviewed the relevant concepts, I'll try to use them to find $\mathbf{b}.$
Since $\mathbf{a}$ is parallel to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$ I can write $\mathbf{a} = k \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ for some constant $k.$
Then, using the given equation $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix},$ I can solve for $\mathbf{b}$ by subtracting $\mathbf{a}$ from both sides.
This gives me
\[\mathbf{b} = \begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix} - \mathbf{a} = \begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix} - k \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.\]
This expression for $\mathbf{b}$ has one unknown variable, $k,$ which I need to find.
To do that, I can use the fact that $\mathbf{b}$ is orthogonal to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.$
This means that $\mathbf{b} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0.$
Substituting the expression for $\mathbf{b}$ into this equation, I get
\[\begin{align*}
\left(\begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix} - k \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} &= 0 \\ 
(6 - k) + (-3 - k) + (-6 - k) &= 0 \\
-3k - 3 &= 0 \\
-3k &= 3 \\
k &= -1.
\end{align*}\]
So, I have found the value of $k$ that makes $\mathbf{a}$ parallel to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $\mathbf{b}$ orthogonal to $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.$
Now I can plug this value into the expression for $\mathbf{b}$ to get the final answer.
This gives me
\[\mathbf{b} = \begin{pmatrix} 6 \\ -3 \\ -6 \end{pmatrix} - (-1) \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 7 \\ -2 \\ -5 \end{pmatrix}.\]