Example 1: Write the equation of a line through a point with a given direction vector
Write the vector equation of the line through \( A(2,-1,3) \) with direction vector
\[
\begin{pmatrix}
4\\
-2\\
1
\end{pmatrix}.
\]
Solution
Using
\[
\mathbf{r}=
\begin{pmatrix}
x_1\\
y_1\\
z_1
\end{pmatrix}
+
\lambda
\begin{pmatrix}
a\\
b\\
c
\end{pmatrix},
\]
we get
\[
\mathbf{r}=
\begin{pmatrix}
2\\
-1\\
3
\end{pmatrix}
+
\lambda
\begin{pmatrix}
4\\
-2\\
1
\end{pmatrix}.
\]
Parametric form:
\[
x=2+4\lambda,
\qquad
y=-1-2\lambda,
\qquad
z=3+\lambda.
\]
\( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form:
\[
\mathbf{r}=(2\mathbf{i}-\mathbf{j}+3\mathbf{k})
+
\lambda(4\mathbf{i}-2\mathbf{j}+\mathbf{k}).
\]
All three forms describe the same line.
Example 2: Find the vector equation of the line through two points
Find the equation of the line through \( A(1,2,-3) \) and \( B(5,-1,7) \).
Solution
First find a direction vector:
\[
\overrightarrow{AB}=
\begin{pmatrix}
5-1\\
-1-2\\
7-(-3)
\end{pmatrix}
=
\begin{pmatrix}
4\\
-3\\
10
\end{pmatrix}.
\]
Now use point \( A \):
\[
\mathbf{r}=
\begin{pmatrix}
1\\
2\\
-3
\end{pmatrix}
+
\lambda
\begin{pmatrix}
4\\
-3\\
10
\end{pmatrix}.
\]
Parametric form:
\[
x=1+4\lambda,
\qquad
y=2-3\lambda,
\qquad
z=-3+10\lambda.
\]
\( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form:
\[
\mathbf{r}=(\mathbf{i}+2\mathbf{j}-3\mathbf{k})
+
\lambda(4\mathbf{i}-3\mathbf{j}+10\mathbf{k}).
\]
A line through two points is found by using one point and the displacement vector between the points.
Example 3: Find whether a point lies on a line
Determine whether the point \( P(7,-4,5) \) lies on the line
\[
\mathbf{r}=
\begin{pmatrix}
1\\
2\\
-1
\end{pmatrix}
+
\lambda
\begin{pmatrix}
2\\
-2\\
3
\end{pmatrix}.
\]
Solution
From the vector equation,
\[
x=1+2\lambda,
\qquad
y=2-2\lambda,
\qquad
z=-1+3\lambda.
\]
Using \( x=7 \):
\[
1+2\lambda=7
\qquad \Rightarrow \qquad
2\lambda=6
\qquad \Rightarrow \qquad
\lambda=3.
\]
Check the other coordinates:
\[
y=2-2(3)=-4,
\qquad
z=-1+3(3)=8.
\]
The \( y \)-coordinate matches, but the \( z \)-coordinate does not match \( 5 \).
So \( P(7,-4,5) \) does not lie on the line.
Example 4: Find the point where \( \lambda=2 \)
For the line
\[
\mathbf{r}=
\begin{pmatrix}
-3\\
1\\
4
\end{pmatrix}
+
\lambda
\begin{pmatrix}
5\\
2\\
-1
\end{pmatrix},
\]
find the coordinates of the point when \( \lambda=2 \).
Solution
\[
\mathbf{r}=
\begin{pmatrix}
-3\\
1\\
4
\end{pmatrix}
+
2
\begin{pmatrix}
5\\
2\\
-1
\end{pmatrix}
=
\begin{pmatrix}
-3\\
1\\
4
\end{pmatrix}
+
\begin{pmatrix}
10\\
4\\
-2
\end{pmatrix}
=
\begin{pmatrix}
7\\
5\\
2
\end{pmatrix}.
\]
The point is \((7,5,2)\).
Example 5: Convert from \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form to parametric form
Convert
\[
\mathbf{r}=(3\mathbf{i}-2\mathbf{j}+\mathbf{k})
+
t(2\mathbf{i}+5\mathbf{j}-4\mathbf{k})
\]
into parametric equations.
Solution
Match the coefficients of \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \):
\[
x=3+2t,
\qquad
y=-2+5t,
\qquad
z=1-4t.
\]
The parametric form is
\[
x=3+2t,\qquad y=-2+5t,\qquad z=1-4t.
\]