Question

Question: Find the parametric equations for the line through the points \(\left( {0,\frac{1}{2},1} \right)\) and \((2,1, - 3)\) and the symmetric equations for the line through the points \(\left( {0,\frac{1}{2},1} \right)\) and \((2,1, - 3)\).

Short answer

The parametric equations for the line through the points \(\left( {0,\frac{1}{2},1} \right)\) and \((2,1, - 3)\) are \(x = 2t,y = 0.5 + 0.5t,z = 1 - 4t\).

The symmetric equations for the line through the points \(\left( {0,\frac{1}{2},1} \right)\) and \((2,1, - 3)\) are \(\frac{x}{2} = \frac{{y - 0.5}}{{0.5}} = \frac{{z - 1}}{{ - 4}}\).

Step-by-step solution

Write the expression to find parametric equations, direction vector \((r)\) and symmetric equations.

Formula used

\(\begin{array}{l}x = {x_0} + at,\\y = {y_0} + bt,\\z = {z_0} + ct(1)\end{array}\)

\(\langle a,b,c\rangle = \left\langle {\left( {{x_1} - {x_0}} \right),\left( {{y_1} - {y_0}} \right),\left( {{z_1} - {z_0}} \right)} \right\rangle (2)\)

\(\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}\)

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

Consider the points \({P_0}\) is \(\left( {0,\frac{1}{2},1} \right)\) and \({P_1}\) is \((2,1, - 3)\).

In equation (2), substitute 0 for \({x_0},\frac{1}{2}\) for \({y_0},1\) for \({z_0},2\) for \({x_1},1\) for \({y_1}\), and \( - 3\) for \({z_1}\).

\(\begin{array}{l}\langle a,b,c\rangle = \left\langle {(2 - 0),\left( {1 - \frac{1}{2}} \right),( - 3 - 1)} \right\rangle \\\langle a,b,c\rangle = \left\langle {2,\frac{1}{2}, - 4} \right\rangle \end{array}\)

Calculate the parametric equations and substitute the values.

In equation (1), substitute 0 for \({x_0},\frac{1}{2}\) for \({y_0},1\) for \({z_0},2\) for \(a,\frac{1}{2}\) for \(b\), and \( - 4\) for \(c\).

\(\begin{array}{l}x = 0 + (2)t,\\y = \frac{1}{2} + \left( {\frac{1}{2}} \right)t,\\z = 1 + ( - 4)t\\x = 2t,\\y = 0.5 + 0.5t,\\z = 1 - 4t\end{array}\)

Thus, the parametric equations for the line through the points \(\left( {0,\frac{1}{2},1} \right)\) and \((2,1, - 3)\) are \(x = 2t,y = 0.5 + 0.5t,z = 1 - 4t\)

In equation (3), substitute 0 for \({x_0},\frac{1}{2}\) for \({y_0},1\) for \({z_0},2\) for \(a,\frac{1}{2}\) for \(b\), and \( - 4\) for \(c\).

\(\begin{array}{l}\frac{{x - 0}}{2} = \frac{{y - \frac{1}{2}}}{{\frac{1}{2}}}\\\frac{{x - 0}}{2} = \frac{{z - 1}}{{ - 4}}\frac{x}{2}\\\frac{{x - 0}}{2} = \frac{{y - 0.5}}{{0.5}}\\\frac{{x - 0}}{2} = \frac{{z - 1}}{{ - 4}}\end{array}\)

As a result, the symmetric equations for the line connecting the points \(\left( {0,\frac{1}{2},1} \right)\) and \((2,1, - 3)\) are \(\underline {\frac{x}{2}} = \frac{{y - 0.5}}{{0.5}} = \frac{{z - 1}}{{ - 4}}\).