Converting the equation of the line to the vector and symmetric forms of the equation.

Question

Does anyone know how to convert the equation of the line to the vector and symmetric forms:

$x=-1-2t$

$y=1+3t$

$z=2+t$

Answer 1

The vector form is given simply rewriting the three equations in vector form: $$ \begin{bmatrix} x\\y\\z \end{bmatrix}= \begin{bmatrix} -1\\1\\2 \end{bmatrix}+t \begin{bmatrix} -2\\3\\1 \end{bmatrix} $$

for the symmetric form find $t$ from the three equations: $$ t=\frac{x+1}{-2} \qquad t=\frac{y-1}{3} \qquad t=z-2 $$ so you have: $$ \frac{x+1}{-2}=\frac{y-1}{3}=z-2 $$

Answer 2

Here's my approach

EQUATION OF LINE IN VECTOR FORM $$ \vec{r} = \vec{a} + \lambda \vec{b} $$

and in SYMMETRIC FORM $$ \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} $$ we can write it also as $$ \frac{x-\left ( \vec{a}\cdot \hat{i}\right )}{\vec{b}\cdot \hat{i}}=\frac{y-\left ( \vec{a}\cdot \hat{j}\right )}{\vec{b}\cdot \hat{j}}=\frac{z-\left ( \vec{a}\cdot \hat{k}\right )}{\vec{b}\cdot \hat{k}} $$

By comparing it with the given data

$$ \\ \frac{x+1}{-2}=\frac{y-1}{3}=\frac{z-2}{1} \\ \vec{a} = -\hat{i} + \hat{j} + 2\hat{k} \\ \vec{b} = -2\hat{i} + 3\hat{j} + \hat{k} $$

ANSWER $$ \\ \vec{r} = -\hat{i} + \hat{j} + 2\hat{k} +\lambda\left (-2\hat{i} + 3\hat{j} + \hat{k} \right ) $$