I am having some issues trying to complete this calculus problem. I know that a vector equation is of the form $$\langle a,b,c\rangle \big(\langle x,y,z\rangle - \langle x_0,y_0,z_0\rangle \big)=0$$ and that parametric are of the form $x = a+vt$, $y = b+vt$, $z = c + vt$. What exactly should I be doing to find that specific vector equation and parametric equation with the given information?
Find a vector equation and parametric equations in t for the line through the point and perpendicular to the given plane. ($P_0$ corresponds to $t = 0$.)
$P_0 = (4, 0, 7)$
$x + 3y + z = 9$
$v = ?i + ?j + ?k$
$x = ?$
$y = ?$
$z = ?$
1 Answer
I think there's a misunderstanding of the parametric equations of a straight line here: $\vec v$, being a vector, can't be found in scalar equations such as $x=a+vt$.
Using the notations of affine geometry, the vector equation will be of the form $P=P_0+t\vec v$, where $\vec v$ is the direction vector of the line.
Now since the straight line has to be perpendicular to the plane with equation $\;x+3y+z=9$, we know a normal vector to the plane is the vector $\vec n=(1,3,1)$, so a vector equation is $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\0\\7\end{pmatrix}+t\begin{pmatrix}1\\3\\1\end{pmatrix}\iff\begin{cases}x=4+t,\\y=3t,\\z=7+t.\end{cases}$$
