How to find parametric equation of the line which is perpendicular to 2 lines and passes through point of intersection?

Question

$$ \left\{\begin{array}{lcccc} \mbox{Line}\ 1 & : & x = 1 + 2a, & y = 2 - a, & z = 4 - 2a \\[1mm] \mbox{Line}\ 2 & : & \!\! x = 9 + b, & \,\,\, y = 5 + 3b, & \,\,z = -4-b \end{array}\right. $$ Point of Intersection: $\left[7,-1,-2\right]$.

How to find parametric equation of the line which is perpendicular to these $2$ lines and passes though point of intersection ?.

Answer 1

Hint:

The coefficient vector of the parameter $a$ represents the "direction vector" of that line, and taking the cross product with another vector will give a perpendicular direction.

Answer 2

Take a plane $\alpha$ passing through the point and perpendicular to line 1. Take a plane $\beta$ passing throught the point and perpendicular to line 2. The intersection $\alpha \cap \beta$ is the line you're looking for.

Answer 3

The parametric equations of lines are:

$L_1$: $\frac {x-1} 2=\frac{y-2}{-1}=\frac{z-4}{-2}=a$

$L_2$: $\frac {x-9} 1=\frac{y-5}{3}=\frac{z+4}{-1}=b$

So the gradients of normal to the plain containing two lines are:

$(l, m, n)=[2\times1=2, 3(-1)=-3, (-2)(-1)=2]$

Therefore the equation of required line is:

$\frac{x-7} 2=\frac{y+1} {-3}=\frac{z+2} 2=t$

You can find x, y and z for t and find similar equation as lines $L_1$ and $L_2$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\left\{\begin{array}{lcl} \mbox{Line}\ 1 & : & \vec{r} = \overbrace{\phantom{A}\vec{r}_{1}\phantom{A}} ^{\ds{\pars{\begin{array}{c}1 \\ 2 \\ 4\end{array}}}}\ + a\ \overbrace{\phantom{A}\vec{n}_{1}\phantom{A}} ^{\ds{\pars{\begin{array}{r}2 \\ -1 \\ -2\end{array}}}} \\[1mm] --- & -- & ------------- \\[1mm] \mbox{Line}\ 2 & : & \vec{r} = \underbrace{\phantom{A}\vec{r}_{2}\phantom{A}} _{\ds{\pars{\begin{array}{r}9 \\ 5 \\ -4\end{array}}}}\ + a\ \underbrace{\phantom{A}\vec{n}_{2}\phantom{A}} _{\ds{\pars{\begin{array}{r}1 \\ 3 \\ -1\end{array}}}} \end{array}\right. \end{align}

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The answer is given by \begin{align} \mbox{} & \\ \bbox[10px,#ffd,border:1px groove navy]{% \pars{\begin{array}{c}x \\ y \\ z\end{array}}} & = \pars{\begin{array}{r}7 \\ -1 \\ -2\end{array}}\ +\ a\,\vec{n}_{1} \times \vec{n}_{2} = \pars{\begin{array}{r}7 \\ -1 \\ -2\end{array}}\ +\ a\verts{\begin{array}{r} \hat{x} & \hat{y} & \hat{z} \\ 2 & -1 & 2 \\ 1 & 3 & -1 \end{array}} = \pars{\begin{array}{r}7 \\ -1 \\ -2\end{array}}\ +\ a\pars{\begin{array}{r}-1 \\ 4 \\ 7\end{array}} \\[5mm] & = \bbox[10px,#ffd,border:1px groove navy]{% \pars{\begin{array}{rcr}7 & - & a\\ -1 & + &4a \\ -2 & + & 7a\end{array}}} \\ & \end{align}