### Instruction: A line is parameterized by a parameter $t,$ so that the vector on the line at $t = -1$ is $\begin{pmatrix} 1 \\ 3 \\ 8 \end{pmatrix},$ and the vector on the line at $t = 2$ is $\begin{pmatrix} 0 \\ -2 \\ -4 \end{pmatrix}.$  Find the vector on the line at $t = 3.$.### Response: I need to find a general formula for the vector on the line as a function of $t.$
Since the line is parameterized by $t,$ I can write the vector on the line as $\begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix},$ where $x, y, z$ are functions of $t.$
I can use the given information to find two equations for each of these functions.
For example, for $x(t),$ I have $x(-1) = 1$ and $x(2) = 0.$
I can assume that $x(t)$ is a linear function of $t,$ since the line is straight.
So I can write $x(t) = at + b,$ where $a$ and $b$ are constants.
To find $a$ and $b,$ I can plug in the values of $t$ and $x(t)$ that I know.
I get $1 = -a + b$ and $0 = 2a + b.$
I can solve this system of equations by subtracting the first from the second, which gives me $-1 = 3a.$
So $a = -\frac{1}{3}.$
Then I can plug in $a$ into either equation to find $b.$
I get $b = 1 - a = 1 + \frac{1}{3} = \frac{4}{3}.$
So $x(t) = -\frac{1}{3}t + \frac{4}{3}.$
I can repeat the same process for $y(t)$ and $z(t),$ using the given values of $y(-1), y(2), z(-1),$ and $z(2).$
For $y(t),$ I get $y(-1) = 3$ and $y(2) = -2.$
So $y(t) = ct + d,$ where $c$ and $d$ are constants.
I get $3 = -c + d$ and $-2 = 2c + d.$
Subtracting, I get $5 = -3c.$
So $c = -\frac{5}{3}.$
Then $d = 3 + c = 3 - \frac{5}{3} = \frac{4}{3}.$
So $y(t) = -\frac{5}{3}t + \frac{4}{3}.$
For $z(t),$ I get $z(-1) = 8$ and $z(2) = -4.$
So $z(t) = et + f,$ where $e$ and $f$ are constants.
I get $8 = -e + f$ and $-4 = 2e + f.$
Subtracting, I get $12 = -3e.$
So $e = -4.$
Then $f = 8 + e = 8 - 4 = 4.$
So $z(t) = -4t + 4.$
Now I have a formula for the vector on the line as a function of $t.$
It is $\begin{pmatrix} -\frac{1}{3}t + \frac{4}{3} \\ -\frac{5}{3}t + \frac{4}{3} \\ -4t + 4 \end{pmatrix}.$
To find the vector on the line at $t = 3,$ I just need to plug in $t = 3$ into this formula.
I get $\begin{pmatrix} -\frac{1}{3}(3) + \frac{4}{3} \\ -\frac{5}{3}(3) + \frac{4}{3} \\ -4(3) + 4 \end{pmatrix}.$
Simplifying, I get $\begin{pmatrix} -1 + \frac{4}{3} \\ -5 + \frac{4}{3} \\ -12 + 4 \end{pmatrix}.$
Combining the fractions, I get $\begin{pmatrix} \frac{1}{3} \\ -\frac{11}{3} \\ -8 \end{pmatrix}.$