Question

Find a vector equation and parametric equations for the line segment that joins p to q.P(0, - 1, 1), Q(1/2, 1/3, 1/4)

Answer 1

Answer:

vector equation:

$\overrightarrow{PQ}=\begin{bmatrix}1/2}\\4/3\\-3/4\end{bmatrix}$

parametric equations:

$r_x = \dfrac{1}{2}t\\r_y = -1 +\dfrac{4}{3}t\\r_z = 1=\dfrac{3}{4}t$

Step-by-step explanation:

The coordinates of the points are given as:

P(0,-1,1) and Q(1/2,1/3,1/4)

the coordinates of any points are also position vectors (vectors starting from the origin to that point), and can be represented as:

$\overrightarrow{OP} = 0\hat{i}-1\hat{j}+1\hat{k}$

or

$\overrightarrow{OP} = \begin{bmatrix}0\\-1\\1\end{bmatrix}$

similarly,

$\overrightarrow{OQ} =\dfrac{1}{2}\hat{i}+\dfrac{1}{3}\hat{j}+\dfrac{1}{4}\hat{k}$

or

$\overrightarrow{OQ} = \begin{bmatrix}1/2}\\1/3\\1/4\end{bmatrix}$

the vector PQ can be described as:

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$

$\overrightarrow{PQ}=\begin{bmatrix}1/2}\\1/3\\1/4\end{bmatrix}-\begin{bmatrix}0\\-1\\1\end{bmatrix}$

$\overrightarrow{PQ}=\begin{bmatrix}1/2}\\4/3\\-3/4\end{bmatrix}$

this is the vector equation of the line segment from P to Q.

to make the parametric equations:

we know that the general equation of a line is represented as:

$\overrightarrow{r} = \overrightarrow{r_0} + t\overrightarrow{d}$

here, $\overrightarrow{r_0}$: is the initial position or the starting point. in our case it is the position vector of P

and $\overrightarrow{d}$: is the direction vector or the direction of the line. in our case that's PQ vector.

$\overrightarrow{r} = \begin{bmatrix}0\\-1\\1\end{bmatrix} + t\begin{bmatrix}1/2}\\4/3\\-3/4\end{bmatrix}$

that parametric equations can now be easily formed:

$\begin{bmatrix}r_x\\r_y\\r_z\end{bmatrix} = \begin{bmatrix}0\\-1\\1\end{bmatrix} + t\begin{bmatrix}1/2}\\4/3\\-3/4\end{bmatrix}$

$r_x = \dfrac{1}{2}t\\r_y = -1 +\dfrac{4}{3}t\\r_z = 1=\dfrac{3}{4}t$

these are the parametric equations of the line PQ