Question

How do you find the intersection of two lines in three-dimensional space?

Answer 1

Hint: To find the intersection of two lines in three dimensional space, if the two given lines are intersected lines or they intersect each other at some specific values of \[\lambda \ and\ \mu \] so that \[\overrightarrow{{{r}_{1}}}=\overrightarrow{{{r}_{2}}}\]. Solving the given equations of lines in 3-D space, we will get the values of \[\lambda \ and\ \mu \]. We will solve the given two lines in a way we solve the general pair of linear equations by substitution method. At last substituting the values of the given parameters, we will get the intersection point.

Complete step by step solution:
To find the intersection of two lines in three dimensional space:
We can understand this by taking an imaginary example.
Suppose we have the two given lines,
\[{{L}_{1}}:\ \overrightarrow{{{r}_{1}}}=\left( \begin{matrix}
   5 \\
   2 \\
   -1 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
   1 \\
   -2 \\
   -3 \\
\end{matrix} \right)\]
\[{{L}_{2}}:\ \overrightarrow{{{r}_{2}}}=\left( \begin{matrix}
   2 \\
   0 \\
   4 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
   1 \\
   2 \\
   -1 \\
\end{matrix} \right)\]
If the two given lines are intersected lines or they intersect each other then for some specific values of \[\lambda \ and\ \mu \], we will get
\[\overrightarrow{{{r}_{1}}}=\overrightarrow{{{r}_{2}}}\]
Therefore,
\[\left( \begin{matrix}
   5 \\
   2 \\
   -1 \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
   1 \\
   -2 \\
   -3 \\
\end{matrix} \right)=\left( \begin{matrix}
   2 \\
   0 \\
   4 \\
\end{matrix} \right)+\mu \left( \begin{matrix}
   1 \\
   2 \\
   -1 \\
\end{matrix} \right)\]
Now,
Comparing the coefficients of\[\hat{i},\ \hat{j}\ and\ \hat{k}\],
We have three equations;
\[\hat{i}:\ \ 5+\lambda =2+\mu \]------ (1)
\[\hat{j}:\ \ 2-2\lambda =2\mu \] Or \[1-1\lambda =\mu \]------ (2)
\[\hat{k}:\ \ -1-3\lambda =4-\mu \]------ (3)
Adding equation (1) and equation (2), we will get
\[\Rightarrow 6=2+2\mu \]
\[\Rightarrow \mu =2\]
Now,
Putting the value of \[\mu =2\]in equation (1),
\[\ \Rightarrow 5+\lambda =2+2\]
\[\Rightarrow 5+\lambda =4\]
\[\Rightarrow \lambda =-1\]
Thus,
We get the values of the two parameters i.e. \[\mu \ and\ \lambda \]
\[\Rightarrow \mu =2\ and\ \lambda =-1\]
Substitution the value of \[\mu =2\ and\ \lambda =-1\] in\[{{L}_{1}}\ and\ {{L}_{2}}\],
\[{{L}_{1}}:\ \overrightarrow{{{r}_{1}}}=\left( \begin{matrix}
   5 \\
   2 \\
   -1 \\
\end{matrix} \right)-1\left( \begin{matrix}
   1 \\
   -2 \\
   -3 \\
\end{matrix} \right)=\left( \begin{matrix}
   5-1 \\
   2+2 \\
   -1+3 \\
\end{matrix} \right)=\left( \begin{matrix}
   4 \\
   4 \\
   2 \\
\end{matrix} \right)\]
\[{{L}_{2}}:\ \overrightarrow{{{r}_{2}}}=\left( \begin{matrix}
   2 \\
   0 \\
   4 \\
\end{matrix} \right)+2\left( \begin{matrix}
   1 \\
   2 \\
   -1 \\
\end{matrix} \right)=\left( \begin{matrix}
   2+2 \\
   4 \\
   4-2 \\
\end{matrix} \right)=\left( \begin{matrix}
   4 \\
   4 \\
   2 \\
\end{matrix} \right)\]
Thus,
We get the coordinate of the point of intersection in three-dimensional space is \[\left( 4,4,2 \right)\].
Therefore,
In this example \[{{L}_{1}}\ and\ {{L}_{2}}\]intersects at a point i.e. \[\left( 4,4,2 \right)\].

Note:
Given two lines in 3-D space, they may satisfy any one of the following conditions;
They will intersect at a point.
They will be parallel to each other and do not intersect each other.
The two lines are parallel and coincident. This means that the given lines are the same and all the points will satisfy both the linear equation.
The two lines cannot intersect as well as not parallel that means they are skew lines.
For the intersection of the two given lines at a point or coincident, the lines should be coplanar lines i.e. they must lie on the same plane. Two lines that are parallel to each other are also the coplanar lines. The skew lines are those lines that do not lie on the same plane.