Question

How do you find parametric equations for the line through ${P}_{0}$=(3, -1, 1) perpendicular to the plane $3x+5y-7z=29$?

Answer 1

For finding the equation of the line, we need the coordinates of the point through which it passes, and a vector parallel to it. The coordinates of the point are given as ${{P}_{0}}=\left( 3,-1,1 \right)$, and the line is given to be normal to the plane $3x+5y-7z=29$, which means that it will be parallel to the normal vector to the plane. The standard equation of a line in three dimensions is given as $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$. On substituting the coordinates and the direction cosine of the normal vector into the standard equation, we will obtain the equation of the line. Finally, on equating each of the terms to a parameter $\lambda$, we will obtain the parametric equations.

Complete step-by-step solution:
We know that in the three dimensional geometry, a line is defined by a point through which it passes and a vector parallel to the given line. If the coordinates of the point through which the line passes are $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, and the vector parallel to the line is given by $\vec{v}=a\hat{i}+b\hat{j}+c\hat{k}$, then its equation can be given by
$\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}........\left( i \right)$
Now, according to the question, the unknown line passes through a point ${{P}_{0}}$ whose coordinates are given as $\left( 3,-1,1 \right)$. So we have got the coordinates of the point through which the line passes. Now, we need the vector parallel to the unknown line. For this, it is given in the question that the line is normal to the plane whose equation is given as
$\Rightarrow 3x+5y-7z=29$
Since the line is normal to the given plane, it must be parallel to the vector normal to the plane. Now, we know that the direction cosines of the normal vector to a plane are given by the coefficients of x, y and z in its equation. Therefore, from the above equation we can write the normal vector to the given plane as
$\Rightarrow \vec{n}=3\hat{i}+5\hat{j}-7\hat{k}$
Therefore, from the equation (i) we can write the equation of the line as
$\begin{aligned} & \Rightarrow \dfrac{x-3}{3}=\dfrac{y-\left( -1 \right)}{5}=\dfrac{z-1}{-7} \\\ & \Rightarrow \dfrac{x-3}{3}=\dfrac{y+1}{5}=\dfrac{z-1}{-7} \\\ \end{aligned}$
For getting the parametric equation of the line, we equate each of the three terms to a parameter, say $\lambda$ so that we can write
$\Rightarrow \dfrac{x-3}{3}=\dfrac{y+1}{5}=\dfrac{z-1}{-7}=\lambda$
From the first equality, we get
$\begin{aligned} & \Rightarrow \dfrac{x-3}{3}=\lambda \\\ & \Rightarrow x=3\lambda +3 \\\ & \Rightarrow x=3\left( \lambda +1 \right) \\\ \end{aligned}$
From the second equality we get
$\begin{aligned} & \Rightarrow \dfrac{y+1}{5}=\lambda \\\ & \Rightarrow y=5\lambda -1 \\\ \end{aligned}$
And from the third equality, we get
$\begin{aligned} & \Rightarrow \dfrac{z-1}{-7}=\lambda \\\ & \Rightarrow z=-7\lambda +1 \\\ \end{aligned}$
Hence, we have obtained the parametric equations of the line as $x=3\left( \lambda +1 \right)$, $y=5\lambda -1$ and $z=-7\lambda +1$.

Note: For solving the question of these types, we must be familiar with the concepts of the three dimensional geometry. The concept of line in three dimensions is different to that in the two dimensions by the fact that the orientation of the line in $3D$ is decided by the parallel vector instead of the slope in $2D$.