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Question: How do you find the set of parametric equations for the line in 3D described by general equations \[...

How do you find the set of parametric equations for the line in 3D described by general equations x−y−z=−4 and x+y−5z=−12x-y-z=-4\ and \ x+y-5z=-12?

Explanation

Solution

in the given question, we have been asked to find a line that constitutes the intersection of two given planes. The equation of a line in 3D form is r→=r0→+λd→\overrightarrow{r}=\overrightarrow{{{r}_{0}}}+\lambda \overrightarrow{d} where d is the direction vector and r0{{r}_{0}} is vector at a point of the line. First we need to find the vector at a point of line then we will evaluate the direction of the vector and put all the values in the equation of a line in 3D form. In this way we will get our required solution.

Complete step by step solution:
We have given that,
x−y−z=−4x-y-z=-4
x+y−5z=−12x+y-5z=-12
Using the equation of a line in 3D, i.e.
r→=r0→+λd→\overrightarrow{r}=\overrightarrow{{{r}_{0}}}+\lambda \overrightarrow{d}
To find the value of ‘x’, ‘y’ and ‘z’,
Putting z = 0 in the above equations, we get
x−y=−4x-y=-4
x+y=−12x+y=-12
Solving the above equations by substitution method, we get
x−y=−4⇒x=−4+yx-y=-4\Rightarrow x=-4+y
Substitute in the other equation, we get
x+y=−12⇒−4+y+y=−12⇒−4+2y=−12⇒y=−4x+y=-12\Rightarrow -4+y+y=-12\Rightarrow -4+2y=-12\Rightarrow y=-4
Substitute y = -4 inx=−4+yx=-4+y, we get
x=−4+y⇒x=−4−4=−8x=-4+y\Rightarrow x=-4-4=-8
∴x=−8 and y=−4\therefore x=-8\ and\ y=-4
Now,
We have the fixed point, i.e.

{{x}_{0}} \\\ {{y}_{0}} \\\ {{z}_{0}} \\\ \end{matrix} \right)=\left( \begin{matrix} -8 \\\ -4 \\\ 0 \\\ \end{matrix} \right)$$ Now, for solving for the direction of the line. We will take x = 0, Substitute the value in the given equation i.e. $$x-y-z=-4\ and\ x+y-5z=-12$$ We have, $$-y-z=-4$$ $$y+5z=-12$$ Adding both the equation, we get $$z=\dfrac{8}{3}$$ Putting the value of $$z=\dfrac{8}{3}$$in any of the above equation, we get $$y=\dfrac{4}{3}$$ Thus, We have the fixed point, i.e. $$\overrightarrow{{{r}_{1}}}=\left( \begin{matrix} {{x}_{1}} \\\ {{y}_{1}} \\\ {{z}_{1}} \\\ \end{matrix} \right)=\left( \begin{matrix} 0 \\\ \left( \dfrac{4}{3} \right) \\\ \left( \dfrac{8}{3} \right) \\\ \end{matrix} \right)$$ Therefore, The direction vector is = $$\overrightarrow{d}=\overrightarrow{{{r}_{1}}}-\overrightarrow{{{r}_{0}}}=\left( \begin{matrix} 0 \\\ \dfrac{4}{3} \\\ \dfrac{8}{3} \\\ \end{matrix} \right)-\left( \begin{matrix} -8 \\\ -4 \\\ 0 \\\ \end{matrix} \right)=\left( \begin{matrix} 8 \\\ \dfrac{16}{3} \\\ \dfrac{8}{3} \\\ \end{matrix} \right)=\dfrac{4}{3}\left( \begin{matrix} 6 \\\ 4 \\\ 2 \\\ \end{matrix} \right)$$ Here we ignore the scalar unit as we only considering the direction, Thus, $$\overrightarrow{d}=\left( \begin{matrix} 6 \\\ 4 \\\ 2 \\\ \end{matrix} \right)$$ Now, Putting all the above values in the equation of liner in 3D, $$\overrightarrow{r}=\overrightarrow{{{r}_{0}}}+\lambda \overrightarrow{d}$$ $$\overrightarrow{r}=\left( \begin{matrix} -8 \\\ -4 \\\ 0 \\\ \end{matrix} \right)+\lambda \left( \begin{matrix} 6 \\\ 4 \\\ 2 \\\ \end{matrix} \right)$$ Therefore, this is the required answer. **Note:** Students need to have the basic knowledge of the concepts of parametric equations in order to solve these types of questions. Students always make sure that they have done correctly while forming the vectors and parametric equations otherwise the vector form that is being formed will be wrong and avoid making errors.