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Vector equation of a line

vector equation of a line

A line in 3-space (or indeed, n-space) can be defined as follows:

Let $A$ be the point such that $OA$ gives the position vector ${\bf a}$,similarly, let $B$ be such that $OB$ has position vector ${\bf b}$, then the equation of the passing through both A and B is $$ {\bf r} = {\bf a} + \lambda({\bf b} - {\bf a}) $$

Here, ${\bf r}$ represents the position vector of an arbitrary point on the line, specified by the parameter $\lambda$

vector equation of a line 1

Determine the vector equation of the line passing through the points $A(1,2,4)$ and $B(2,8,7)$.

Hence determine if the point $C(5,26,16)$ lies on the line. Show also that the point $D(5,12,18)$ does not lie on the line.

solution - press button to display

The equation of the line is $$ {\bf r} = \left(\begin{array}{c}1 \\ 2 \\ 4\end{array}\right) +\lambda\left(\begin{array}{c}1 \\ 6 \\ 3\end{array}\right) $$

To determine if the line passes through $C(5,26,16)$, consider the position vector $OC$. If $C$ lines on the line then, for some value of $\lambda$, $$ OC = \left(\begin{array}{c}1 \\ 2 \\ 4\end{array}\right) +\lambda\left(\begin{array}{c}1 \\ 6 \\ 3\end{array}\right) $$

This gives us three equations, $5 = 1 + \lambda,\; 26 = 2 + 6\lambda,\; 16 = 4 + 3\lambda$, the solution of all three yields $\lambda = 4$, consequently, the line passes through the point $C$. 

In contrast, with the point $D$, we get the following equation $$ OD = \left(\begin{array}{c}1 \\ 2 \\ 4\end{array}\right) +\lambda\left(\begin{array}{c}1 \\ 6 \\ 3\end{array}\right) $$ which gives rise to the three equations $$ 5 = 1 + \lambda, \; 12 = 2 + 6\lambda, \; 18 = 4 + 3\lambda $$ these equations are inconsistent and so we can conclude that the line does not pass through the point $D$.

vector equation of a line 2

Determine the vector equation of the line passing through the points $A(1,5,9)$ and $B(3,-2,6)$.

Hence determine the Cartesian equation of the line.

solution - press button to display

The vector equation of the line is $$ {\bf r} = \left(\begin{array}{c} 1 \\ 5 \\ 9 \end{array}\right) + \lambda \left(\begin{array}{c} 2 \\ -7 \\ -3 \end{array}\right) $$

To find the Cartesian equation, let ${\bf r}$ be given by $${\bf r} = \left(\begin{array}{c} x \\ y \\ z \end{array}\right)$$ then we can get three equations, $x = 1 + 2\lambda,\;y = 5 - 7\lambda,\;z = 9 - 3\lambda$. Making $\lambda$ the subject of these equations gives $$ \lambda = \frac{x-1}{2} = \frac{y-5}{-7} = \frac{z-9}{-3} $$

vector equation of a line 3

A line has vector equation $$ {\bf r} = \left(\begin{array}{c}2 \\ 8\\ 3\end{array}\right) + \lambda \left(\begin{array}{c}1 \\ 4\\ 1\end{array}\right) $$

Determine the minimum distance between the line and the origin

solution - press button to display

Let $P$ by the point on the line closest to the origin. Then $OP$ is perpendicular to the line. This implies that $$OP \cdot \left(\begin{array}{c}1 \\ 4\\ 1\end{array}\right)=0$$

We also note that $P$ is on the line, so there exists $\lambda_p$ such that $$ OP = \left(\begin{array}{c}2 \\ 8\\ 3\end{array}\right) + \lambda_p \left(\begin{array}{c}1 \\ 4\\ 1\end{array}\right) $$

Applying these two facts gives $$ OP\cdot \left(\begin{array}{c}1 \\ 4\\ 1\end{array}\right) = 0 \rightarrow 36 + 18\lambda_p = 0 $$ hence $\lambda_p = -2$

Subsitution of the value of $\lambda_p$ gives $OP = \left(\begin{array}{c}-1 \\ 0\\ 1\end{array}\right)$

The length of $OP$, $|OP| = \sqrt{2}$ and so the minimum distance between the origin and the line is $\sqrt{2}$