Question

Find the symmetric equations $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{6}\mathbf{)}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{7}\mathbf{)}$and the parametric equations $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{)}$ of a line, and/or the equation $\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{10}\mathbf{)}$ of the plane satisfying the following given conditions.

Line through$\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}$ and parallel to $\mathit{i}\mathbf{-}\mathbf{2}\mathit{k}$.

Short answer

The symmetric equations of the line is $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{z}-3}{-2},\mathrm{y}=-1.$

The parametric equation is$\mathrm{r}=(4\mathrm{i}-\mathrm{j}+3\mathrm{k})+(\mathrm{i}-2\mathrm{k})\mathrm{t}.$

Step-by-step solution

Concept of the symmetric and parametric equations.

The symmetric equations of the line passing through $\mathbf{(}{\mathbf{x}}_{\mathbf{0}}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}\mathbf{,}{\mathbf{z}}_{\mathbf{0}}\mathbf{)}$and parallel to $\mathbf{ai}\mathbf{+}\mathbf{bj}\mathbf{+}\mathbf{ck}\mathbf{}$is$\frac{\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{0}}}{\mathbf{a}}\mathbf{,}\frac{\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{0}}}{\mathbf{b}}\mathbf{,}\frac{\mathbf{z}\mathbf{-}{\mathbf{z}}_{\mathbf{0}}}{\mathbf{c}}\mathbf{;}$

The parametric equations of the line are$\mathbf{r}\mathbf{=}\mathbf{(}{\mathbf{x}}_{\mathbf{0}}\mathbf{+}\mathbf{at}\mathbf{,}{\mathbf{y}}_{\mathbf{0}}\mathbf{+}\mathbf{bt}\mathbf{,}{\mathbf{z}}_{\mathbf{0}}\mathbf{+}\mathbf{ct}\mathbf{)}$.

Determine the symmetric equation of a straight line.

The given point is $(4,-1,3)$and the equation is $\mathrm{i}-2\mathrm{k}$.

The symmetric equations of the straight line passing through $(4,-1,3)$and parallel to $\mathrm{i}-2\mathrm{k}$is given by

$\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{a}},\frac{\mathrm{y}-{\mathrm{y}}_0}{\mathrm{b}},\frac{\mathrm{z}-{\mathrm{z}}_0}{\mathrm{c}}$

$\Rightarrow \frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}-(-1)}{0},\frac{\mathrm{z}-3}{-2}$

$\Rightarrow \frac{\mathrm{x}-1}{1}=\frac{\mathrm{z}-3}{-2},\mathrm{y}=-1$

Thus, the required solution is$\Rightarrow \frac{\mathrm{x}-1}{1}=\frac{\mathrm{z}-3}{-2},\mathrm{y}=-1$.

Determine the parametric equation of a straight line.

The symmetric equations of the straight line passing through $(4,-1,3)$and parallel to $\mathrm{i}-2\mathrm{k}$is given by

$\mathrm{x}=4+\mathrm{t}$

$\mathrm{y}=-1+\left(0\right)\mathrm{t}$

$\mathrm{z}=3-2\mathrm{t}$

Or

$\mathrm{r}=(4\mathrm{i}-\mathrm{j}+3\mathrm{k})+(\mathrm{i}-2\mathrm{k})\mathrm{t}.$

Thus, the required solution is$\mathrm{r}=(4\mathrm{i}-\mathrm{j}+3\mathrm{k})+(\mathrm{i}-2\mathrm{k})\mathrm{t}.$