Question

The parametric equations and the symmetric equations for the given line.

Short answer

The parametric equations and the symmetric equations for the given line are \(x = 1 + t,y = - 1 + 2t,z = 1 + t\) and \(x - 1 = \frac{{y + 1}}{2} = z - 3.\)

Step-by-step solution

Write the expression to find parametric equations.

Formula used

1. The parametric equations for a line through the point\(\left( {{x_0},{y_0},{z_0}} \right)\)and parallel to the direction vector\(\langle a,b,c\rangle \)are\(x = {x_0} + at,y = {y_0} + bt,z = {z_0} + ct\)

2. The parametric equations for a line through the point\(\left( {{x_0},{y_0},{z_0}} \right)\)and parallel to the direction vector\(\langle a,b,c\rangle \)are\(\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\).

Calculate the direction vector \((v)\). 

The given point is \((1, - 1,1)\) and the line equation is \(x + 2 = \frac{1}{2}y = z - 3\).

Rewrite the line equation .

From the above equation, the direction vector \((v)\) is \(\langle a,b,c\rangle = \langle 1,2,1\rangle \)

Calculate the parametric equations and Symmetric equations.

Substitute \({x_0} = 1,{y_0} = - 1,{z_0} = 1,a = 1,b = 2\), and \(c = 1\) in the formula (1).

\(\begin{array}{l}x = 1 + (1)t,y = - 1 + (2)t,z = 1 + (1)t\\x = 1 + t,y = - 1 + 2t,z = 1 + t\end{array}\)

Thus, the parametric equation for the line through the point \((1, - 1,1)\) and parallel to the line \(x + 2 = \frac{1}{2}y = z - 3\) are \(x = 1 + t,y = - 1 + 2t,z = 1 + t\), respectively.

For symmetric equations

Substitute \({x_0} = 1,{y_0} = - 1,{z_0} = 1,a = 1,b = 2\), and \(c = 1\) in the formula (2).

\(\begin{array}{l}\frac{{x - 1}}{1} = \frac{{y - ( - 1)}}{2} = \frac{{z - 1}}{1}\\x - 1 = \frac{{y + 1}}{2} = z - 3\end{array}\)

Thus, the symmetric equations for the line through the point \((1, - 1,1)\) and parallel to the line \(x + 2 = \frac{1}{2}y = z - 3\) are \(x - 1 = \frac{{y + 1}}{2} = z - 3\).