Find equation of the line containing the given point (5,3,4) and parallel to the vector
-> -> ->
2 i + 5 j - 8 k
I fear that one either understands or not: it is that simple.
$\displaystyle \begin{gathered}
\ell :\left\langle {5,3,4} \right\rangle + t\left\langle {2,5, - 8} \right\rangle \hfill \\
\ell :\left\langle {5 + 2t,3 + 5t,4 - 8t} \right\rangle \hfill \\
\end{gathered} $
$\displaystyle \ell :\left\{ \begin{gathered}
x = 5 + 2t \hfill \\
y = 3 + 5t \hfill \\
z = 4 - 8t \hfill \\
\end{gathered} \right.$
Hello, Faz!
You're expected to know the formulas for this problem.
Find equation of the line containing the given point (5,3,4)
and parallel to the vector $\displaystyle \vec c \:=\:2\vec i + 5\vec j - 8\vec k$
Parametric equations: .$\displaystyle \begin{array}{ccc}x &=&5+2t \\ y&=&3+5t \\ z&=&4-8t \end{array}$
Symmetric equation: .$\displaystyle \frac{x-5}{2} \:=\:\frac{y-3}{5} \:=\:\frac{z-4}{-8} $
You set your x,y,z so that they start at (5,3,4). These are some kind of x,y,z-intercept if you want.
Then you know that in time t, x is going to increase by 2, y is going to increase by 5 and z is going to increase by -8.
You thus have "3 equations".
$\displaystyle
\ell :\left\{ \begin{gathered}
x = 5 + 2t \hfill \\
y = 3 + 5t \hfill \\
z = 4 - 8t \hfill \\
\end{gathered} \right.
$