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

vector equation:

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

parametric equations:

`r_x = (1)/(2)t\\r_y = -1 +(4)/(3)t\\r_z = 1=(3)/(4)t`

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} =(1)/(2)\hat{i}+(1)/(3)\hat{j}+(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 = (1)/(2)t\\r_y = -1 +(4)/(3)t\\r_z = 1=(3)/(4)t`

these are the parametric equations of the line PQ