# Find a vector parametrization equation \vec{r}(t) for the line through the points P=(0, 0, 5) and...

## Question:

Find a vector parametrization equation {eq}\vec{r}(t) {/eq} for the line through the points {eq}P=(0, 0, 5) {/eq} and {eq}Q=(-2, -1, 2) {/eq} for each of the given conditions on the parameter {eq}t {/eq}.

A) If {eq}\vec{r}(0)=(0, 0, 5) {/eq} and {eq}\vec{r}(7)=(-2, -1, 2) {/eq}.

B) If {eq}\vec{r}(5)=P {/eq} and {eq}\vec{r}(8)=Q {/eq}.

## Parametric Equation for a Line:

Normally, we have one equation for a line in the Cartesian plane, up to scalar multiples of the equation. However, when writing an equation for a line using a parameter, we can specify which parameter value represents a certain point. Because of this, we can have multiple representations of the same line, but we can have different endpoints and a different orientation.

## Answer and Explanation:

Recall that a vector equation for a line is given by $$\vec{r}(t) = \vec{a}t + \vec{b} $$ where {eq}\vec{a} {/eq} is a vector parallel to the line, {eq}\vec{b} {/eq} is a vector representing a point on the line and {eq}t {/eq} is some parameter.

(A) Since {eq}\vec{r}(0) = (0, 0, 5) {/eq}, we can use the equation above to get $$\begin{align*} \vec{r}(0) &= (0, 0, 5) \\ \vec{a}(0) + \vec{b} &= (0, 0, 5) \\ \vec{b} &= (0, 0, 5) \end{align*} $$

Additionally, since {eq}\vec{r}(7) = (-2, -1, 2) {/eq}, we can do the same thing and get $$\begin{align*} \vec{r}(7) &= (-2, -1, 2) \\ \vec{a}(7) + \vec{b} &= (-2, -1, 2) \\ 7\vec{a} + \vec{b} &= (-2, -1, 2) \end{align*} $$ but from the calculation above, we know what {eq}\vec{b} {/eq} is so we can sub that into this equation to solve for {eq}\vec{a} {/eq}. We have $$\begin{align*} 7\vec{a} + \vec{b} &= (-2, -1, 2) \\ 7\vec{a} + (0, 0, 5) &= (-2, -1, 2) \\ 7\vec{a} &= (-2, -1, -3) \\ \vec{a} &= \left( -\frac{2}{7}, -\frac{1}{7}, -\frac{3}{7} \right) \end{align*} $$ so the vector parametrization equation for this line is given by $$\boxed{\vec{r}(t) = \left( -\frac{2}{7}, -\frac{1}{7}, -\frac{3}{7} \right)t + (0, 0, 5)} $$

(B) We can do a similar approach. Using {eq}P = (0, 0, 5) {/eq} and {eq}t = 5 {/eq}, we have $$\begin{align*} \vec{r}(5) &= P \\ 5\vec{a} + \vec{b} &= (0, 0, 5) \qquad (1) \end{align*} $$ Using {eq}Q = (-2, -1, 2) {/eq} and {eq}t = 5 {/eq}, we get $$\begin{align*} \vec{r}(8) &= Q \\ 8\vec{a} + \vec{b} &= (-2, -1, 2) \qquad (2) \end{align*} $$ If we take {eq}(2) {/eq} and subtract {eq}(1) {/eq}, we get $$\begin{align*} (8 \vec{a} + \vec{b}) - (5\vec{a} + \vec{b}) &= (-2, -1, 2) - (0, 0, 5) \\ 3\vec{a} &= (-2, -1, -3) \\ \vec{a} &= \left( -\frac{2}{3}, -\frac{1}{3}, -1\right) \end{align*} $$ Using this fact and {eq}(1) {/eq}, we can solve for {eq}\vec{b} {/eq} as $$\begin{align*} 5\vec{a} + \vec{b} &= (0, 0, 5) \\ 5\left( -\frac{2}{3}, -\frac{1}{3}, -1\right) + \vec{b} &= (0, 0, 5) \\ \vec{b} &= (0, 0, 5) - \left( -\frac{10}{3}, -\frac{5}{3}, -5 \right) \\ \vec{b} &= \left( \frac{10}{3}, \frac{5}{3}, 10 \right) \end{align*} $$ and so the vector parametrization equation for this line is given by $$\boxed{\vec{r}(t) = \left( -\frac{2}{3}, -\frac{1}{3}, -1\right)t + \left( \frac{10}{3}, \frac{5}{3}, 10 \right)} $$