Converting vector-form to cartesian-form line equations

Vector form

A line in vector form has the equation:

\vec r = \vec a + \lambda \vec d

…where:

• \vec r is the position vector of any point on the line,
• \vec a is the position vector of a specific point on the line,
• \vec d is the direction vector of the line,
• \lambda is a scalar multiplier (changing it gives a new point on the line).

See vector line equation.

Cartesian form

Cartesian form is the standard equation of a line (see cartesian line equation, written as:

y = mx + c

(or one of its variants).

Gradient

We can find the gradient of the line from the \vec d vector of the vector form equation.

• The top value of the direction vector \vec d is the change in x.
• The bottom value of the direction vector \vec d is the change in y.

From this, we can use the formula for gradient:

\text{gradient} = \frac{\text{change in } y}{\text{change in } x}

So the gradient of a line with direction vector \vec d=\begin{pmatrix}2\\6\end{pmatrix}, for example, will have gradient \frac62=3

Converting from vector to Cartesian form

Once we have the gradient, we can substitute it into the Cartesian form equation (y = mx + c) to find c.

Convert the vector equation \vec r = \begin{pmatrix}1 \\ 2\end{pmatrix} + \lambda \begin{pmatrix}3 \\ 4\end{pmatrix} to cartesian form

• The direction vector is \vec d = \begin{pmatrix}3 \\ 4\end{pmatrix}
  • Change in x = 3
  • Change in y = 4
• Gradient m = \frac{4}{3}
• \vec a = \begin{pmatrix}1 \\ 2\end{pmatrix} gives us that (1, 2) is a point on the line.
• Substitute into y = mx + c to find c:
  • 2 = \frac{4}{3}(1) + c
  • 2 = \frac{4}{3} + c
  • c = 2 - \frac{4}{3}
  • c = \frac{2}{3} - Answer: y = \frac{4}{3}x + \frac{2}{3}

Convert the vector equation \vec r = \begin{pmatrix}2 \\ 5\end{pmatrix} + \lambda \begin{pmatrix}-1 \\ 2\end{pmatrix} to cartesian form

• The direction vector is \vec d = \begin{pmatrix}-1 \\ 2\end{pmatrix}
  • Change in x = -1
  • Change in y = 2
• Gradient m = \frac{2}{-1} = -2
• \vec a = \begin{pmatrix}2 \\ 5\end{pmatrix} gives us that (2, 5) is a point on the line.
• Substitute into y = mx + c to find c:
  • 5 = -2(2) + c
  • 5 = -4 + c
  • c = 5 + 4
  • c = 9
• Answer: y = -2x + 9

Convert the vector equation \vec r = \begin{pmatrix}-1 \\ 4\end{pmatrix} + \lambda \begin{pmatrix}2 \\ -3\end{pmatrix} to cartesian form

• The direction vector is \vec d = \begin{pmatrix}2 \\ -3\end{pmatrix}
  • Change in x = 2
  • Change in y = -3
• Gradient m = \frac{-3}{2}
• \vec a = \begin{pmatrix}-1 \\ 4\end{pmatrix} gives us that (-1, 4) is a point on the line.
• Substitute into y = mx + c to find c:
  • 4 = \frac{-3}{2}(-1) + c
  • 4 = \frac{3}{2} + c
  • c = 4 - \frac{3}{2}
  • c = \frac{5}{2}
• Answer: y = \frac{-3}{2}x + \frac{5}{2}

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