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

Hint: 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.