We want to find the symmetric equations for L where:
![{\displaystyle L={\frac {x-x_{0}}{a}}={\frac {y-y_{0}}{b}}={\frac {z-z_{0}}{c}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d5b35ee5ffdc78dd3c7a0dde2325b27147e3eb4e)
We can find the constants in the above equation by arranging the vector parametric equation given in the question into the form of:
![{\displaystyle r(t)=\langle x_{0},y_{0},z_{0}\rangle +t\langle a,b,c\rangle }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3356d94711e4e967e6c0f5c571f3f2b4f4cf8980)
Expanding the equation given, we get:
![{\displaystyle r(t)=2\mathbf {i} +3t\mathbf {i} +0\mathbf {j} +4t\mathbf {j} +-1\mathbf {k} +0t\mathbf {k} }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7e8183f4a67e86431e960e0887f05237b2fc400b)
Rearranging:
![{\displaystyle r(t)=(2\mathbf {i} +0\mathbf {j} -1\mathbf {k} )+t(3\mathbf {i} +4\mathbf {j} +0\mathbf {k} ):}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5f66907e3410d6e3d7d185789aab48087960e49b)
Which is simply:
![{\displaystyle r(t)=\langle 2,0,-1\rangle +t\langle 3,4,0\rangle }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9c6cb2aa7e8eac7f709d4f3da2a33cbf921b6517)
where
![{\displaystyle \displaystyle x_{0}=2,y_{0}=0,z_{0}=-1,a=3,b=4,c=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0b3d47b725cf90668b4693ed218e8551621020e4)
Plugging them into the symmetric equation, we get:
![{\displaystyle L={\frac {x-2}{3}}={\frac {y-0}{4}}={\frac {z-(-1)}{0}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f68bffa1f957a09d89c4a162696f23bdb05367c4)
Because we cannot have zero in the denominator, our final answer is:
![{\displaystyle {\color {blue}L={\frac {x-2}{3}}={\frac {y}{4}},\quad z=-1.}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2424a8d6cccc93a02eb71a245e9d96f6b42b900d)