Question

How do you 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.) =$\left( {5,0,8} \right)$,$x + 2y + z = 9$?

Answer 1

In this question we have to find the parametric and vector equations for the given line and plane, If ${r_0}$ is the position vector of the point, then the line must have the form
$\overrightarrow r = {r_0} + tv$,
This is the vector equation of a line in three dimensions. By letting $\overrightarrow r = {r_0} + tv$, ${r_0} = \left( {{x_0},{y_0},{z_0}} \right)$, and $v = \left( {a,b,c} \right)$ we obtain the equation the parametric equations of the line passing through the point ${P_0} = \left( {{x_0},{y_0},{z_0}} \right)$ and parallel to the vector $v = \left( {a,b,c} \right)$:
$x = {x_0} + ta$,$y = {y_0} + tb$,$z = {z_0} + tc$. Now substituting the given values we will get the required equations.

Complete step-by-step solution:
Given point is $\left( {5,0,8} \right)$ and the plane is $x + 2y + z = 9$.
A vector perpendicular to the plane $ax + by + cz + d = 0$ is given by $\left( {a,b,c} \right)$.
So the vector perpendicular to the plane $x + 2y + z = 9$ is $\left( {1,2,1} \right)$,
A line is determined by a point and a direction. Thus, to find an equation representing a line in three dimensions choose a point ${P_0}$ on the line and a non-zero vector $v$ parallel to the line. Since any constant multiple of a vector still points in the same direction, it seems reasonable that a point on the line can be found by starting at the point ${P_0}$ on the line and following a constant multiple of the vector $v$.
By Definition the normal vector is always perpendicular to its plane.
Now the parametric equation of a line through$\left( {{x_0},{y_0},{z_0}} \right)$and parallel to the vector $\left( {a,b,c} \right)$ is
$x = {x_0} + ta$,
$y = {y_0} + tb$,
$z = {z_0} + tc$,
So using the above,
Here ${x_0} = 5,a = 1$, ${y_0} = 0,b = 2$, ${z_0} = 8,c = 1$,
Now substituting the values we get,
So the parametric equation of our line is,
$x = 5 + t$,
$y = 0 + 2t$,$\Rightarrow y = 2t$,
$z = 8 + t$,
Now the vector form of the line is given by, $\overrightarrow r = {r_0} + tv$,
Here ${r_0} = 5\overrightarrow i + 8\overrightarrow k$, and $v = \overrightarrow i + 2\overrightarrow j + \overrightarrow k$,
Now substituting the values we get,
$\overrightarrow r = \left( {5\overrightarrow i + 8\overrightarrow k } \right) + t\left( {\overrightarrow i + 2\overrightarrow j + \overrightarrow k } \right)$,
By simplifying we get,
$\overrightarrow r = \left( {5 + t} \right)\overrightarrow i + \left( {0 + 2t} \right)\overrightarrow j + \left( {8 + t} \right)\overrightarrow k$,
Now the vector equation can also be written as,
$\overrightarrow r = \left( {5,0,8} \right) + t\left( {1,2,1} \right)$,
So the required parametric equation is
$x = 5 + t$,
$y = 0 + 2t$,$\Rightarrow y = 2t$,
$z = 8 + t$, and
The vector equation is $\overrightarrow r = \left( {5,0,8} \right) + t\left( {1,2,1} \right)$.

$\therefore$The vector equation and parametric equations in t for the line through the point and perpendicular to the given $\left( {5,0,8} \right)$,$x + 2y + z = 9$, are
$x = 5 + t$,
$y = 0 + 2t$,$\Rightarrow y = 2t$,
$z = 8 + t$, and
The vector equation is $\overrightarrow r = \left( {5,0,8} \right) + t\left( {1,2,1} \right)$.

Note: The equation of a line in two dimensions is $ax + by = c$; it is reasonable to expect that a line in three dimensions is given by $ax + by + cz + d = 0$;, it turns out that this is the equation of a plane.
A plane does not have an obvious "direction'' as does a line. It is possible to associate a plane with a direction in a very useful way, however: there are exactly two directions perpendicular to a plane. Any vector with one of these two directions is called normal to the plane. So while there are many normal vectors to a given plane, they are all parallel or antiparallel to each other.