Question

How do you find the parametric equation and symmetric equations for the line through ${{t}_{0}}$ and parallel to the given line ${{t}_{0}}=\left( 4,-2,4 \right)$ and $x+1=\dfrac{y}{2}=z+5$?

Answer 1

This type of problem is based on the concept of equation of a line. We know that parallel line equations have the same directional cosines. From the given parallel equation, we find that the directional cosines are (1,2,1). We then have to express the equation line in the form $\dfrac{x-{{x}_{0}}}{l}=\dfrac{y-{{y}_{0}}}{m}=\dfrac{z-{{z}_{0}}}{n}$. Here, l=1, m=2 and n=1. Also we know that the line passes through ${{t}_{0}}=\left( 4,-2,4 \right)$. Therefore, ${{x}_{0}}=4$, ${{y}_{0}}=-2$ and ${{z}_{0}}=4$. Substitute these values to find the symmetric equation. We have to assume the given equation to be ‘t’, that is $\dfrac{x-4}{1}=\dfrac{y+2}{2}=\dfrac{z-4}{1}=t$. Find the value of x, y and z with respect to t which is the parametric equation.

Complete step by step solution:
According to the question, we are asked to find the parametric equation and symmetric equation.
We have been given that the line passes through ${{t}_{0}}=\left( 4,-2,4 \right)$ and is parallel to the line of equation $x+1=\dfrac{y}{2}=z+5$. ----------(1)
We can express equation (1) as
$\dfrac{x+1}{1}=\dfrac{y}{2}=\dfrac{z+5}{1}$ ---------(2)
Here, from equation (2) we get, the directional cosines are (1,2,1).
Here, l=1 which is the directional cosine of x, m=2 which is the directional cosine of y and n=1 which is the directional cosine of z.
The symmetric form of an equation is
$\dfrac{x-{{x}_{0}}}{l}=\dfrac{y-{{y}_{0}}}{m}=\dfrac{z-{{z}_{0}}}{n}$,
Where $\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$is a point in the equation and l, m and n are the directional cosines of the equation.
We know that directional cosines of a parallel equation of line are equal.
Therefore, comparing with the directional cosines of equation (2), we get
$\dfrac{x-{{x}_{0}}}{1}=\dfrac{y-{{y}_{0}}}{2}=\dfrac{z-{{z}_{0}}}{1}$
We can further simplify the equation of line as
$x-{{x}_{0}}=\dfrac{y-{{y}_{0}}}{2}=z-{{z}_{0}}$. --------(3)
We have been given that the equation passes through the point ${{t}_{0}}$, where ${{t}_{0}}=\left( 4,-2,4 \right)$.
We have also assumed $\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)$ to be a point in the equation of line.
Let us equate these two points.
We get ${{x}_{0}}=4$, ${{y}_{0}}=-2$ and ${{z}_{0}}=4$.
On substituting the values in equation (3), we get
$x-4=\dfrac{y-\left( -2 \right)}{2}=z-4$
On further simplifications, we get
$x-4=\dfrac{y+2}{2}=z-4$
Therefore, the symmetric form of the equation of line is $x-4=\dfrac{y+2}{2}=z-4$.
Now, we need to find the parametric equation.
Let us substitute the symmetric equation of line to be equal to t.
$\Rightarrow x-4=\dfrac{y+2}{2}=z-4=t$
We now need to find the value of x, y and z with respect to ‘t’ to find the parametric equation of line.
First consider $x-4=t$.
Let us add 4 on both the sides of the equation.
$\Rightarrow x-4+4=t+4$
We know that terms with the same magnitude and opposite signs cancel out.
Therefore, we get
$x=t+4$
Now, consider $\dfrac{y+2}{2}=t$.
Multiply the whole equation by 2. We get
$\dfrac{y+2}{2}\times 2=t\times 2$
$\Rightarrow \dfrac{y+2}{2}\times 2=2t$
We find that 2 are common in both the numerator and denominator of the LHS. On cancelling 2, we get
$y+2=2t$
Now, we need to subtract 2 from both the sides of the equation. We get
$\Rightarrow y+2-2=2t-2$
We know that terms with the same magnitude and opposite signs cancel out.
Therefore, we get
$y=2t-2$
Let us take 2 common terms in RHS. We get
$y=2\left( t-1 \right)$
Then, consider $z-4=t$.
Let us add 4 on both the sides of the equation.
$\Rightarrow z-4+4=t+4$
We know that terms with the same magnitude and opposite signs cancel out.
Therefore, we get
$z=t+4$
The parametric form of an equation of line is (x,y,z).
Here, $x=t+4$, $y=2\left( t-1 \right)$ and $z=t+4$.
Therefore, the parametric form of the equation of line is $\left( t+4,2\left( t-1 \right),t+4 \right)$.
Hence, the symmetric and parametric form of the equation of line is $x-4=\dfrac{y+2}{2}=z-4$ and $\left( t+4,2\left( t-1 \right),t+4 \right)$ respectively.

Note: Whenever you get this type of problems, we should always know the formula for finding the parametric equations and symmetric equations. Avoid calculation mistakes based on sign conventions. It is advisable to find the parametric equation with respect to t.