Vector line equation

As well as being able to represent a line in cartesian form (e.g. y = mx + c), we can also represent a line using vectors.

Equation of a line in vector form

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

Finding the vectors between points

The first step to finding the equation of a line using vectors is to find the vectors between all points.

Find the vector between the point A=(-2,4) and B=(3,7)

• \vec a=\begin{pmatrix}-2\\4\end{pmatrix}
• \vec b=\begin{pmatrix}3\\7\end{pmatrix}
• \vec{AB}=\vec b - \vec a
  • =\begin{pmatrix}3\\7\end{pmatrix} - \begin{pmatrix}-2\\4\end{pmatrix}
  • =\begin{pmatrix}5\\3\end{pmatrix}
• Answer: \vec{AB}=\begin{pmatrix}5\\3\end{pmatrix}

Find ALL vectors between the points A=(1,2,3), B=(4,0,5) and C=(2,6,1)

• \vec a=\begin{pmatrix}1\\2\\3\end{pmatrix}
• \vec b=\begin{pmatrix}4\\0\\5\end{pmatrix}
• \vec c=\begin{pmatrix}2\\6\\1\end{pmatrix}
• \vec{AB}=\vec b - \vec a
  • =\begin{pmatrix}4\\0\\5\end{pmatrix} - \begin{pmatrix}1\\2\\3\end{pmatrix}
  • =\begin{pmatrix}3\\-2\\2\end{pmatrix}
• \vec{AC}=\vec c - \vec a
  • =\begin{pmatrix}2\\6\\1\end{pmatrix} - \begin{pmatrix}1\\2\\3\end{pmatrix}
  • =\begin{pmatrix}1\\4\\-2\end{pmatrix}
• \vec{BC}=\vec c - \vec b
  • =\begin{pmatrix}2\\6\\1\end{pmatrix} - \begin{pmatrix}4\\0\\5\end{pmatrix}
  • =\begin{pmatrix}-2\\6\\-4\end{pmatrix}
• Answers:
  • \vec{AB}=\begin{pmatrix}3\\-2\\2\end{pmatrix}
  • \vec{AC}=\begin{pmatrix}1\\4\\-2\end{pmatrix}
  • \vec{BC}=\begin{pmatrix}-2\\6\\-4\end{pmatrix}

Finding the vector equation from points

Find a vector equation of the line between the points (2,3) and (5,1)

• Let \vec a be the position vector of A from the origin:
  • \vec A = \begin{pmatrix}2 \\ 3\end{pmatrix}
• \vec{AB} = \vec B - \vec A
  • = \begin{pmatrix}5 \\ 1\end{pmatrix} - \begin{pmatrix}2 \\ 3\end{pmatrix}
  • = \begin{pmatrix}3 \\ -2\end{pmatrix}
• So the direction vector is \begin{pmatrix}3 \\ -2\end{pmatrix}.
• Equation of a line:
  • \vec r = \vec a + \lambda \vec d
  • = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -2 \end{pmatrix}
• Answer: \vec r = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -2 \end{pmatrix}

Note: there are multiple solutions to this line equation from the points given. We can find other ones by using different vectors (e.g. using \vec{OB} instead of \vec{OA}).

Find a vector equation of the line between the points (1,0,2) and (4,6,5)

• Let \vec a be the position vector of A from the origin:
  • \vec A = \begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}
• \vec{AB} = \vec B - \vec A
  • = \begin{pmatrix}4 \\ 6 \\ 5\end{pmatrix} - \begin{pmatrix}1 \\ 0 \\ 2\end{pmatrix}
  • = \begin{pmatrix}3 \\ 6 \\ 3\end{pmatrix}
• So the direction vector is \begin{pmatrix}3 \\ 6 \\ 3\end{pmatrix}, which we can simplify to \begin{pmatrix}1 \\ 2 \\ 1\end{pmatrix} (because it’s just a direction and the magnitude is not important here).
• Equation of a line:
  • \vec r = \vec a + \lambda \vec d
  • = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}
• Answer: \vec r = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}

Note: there are multiple solutions to this line equation from the points given. We can find other ones by using different vectors (e.g. using \vec{OB} instead of \vec{OA}).

Find a vector equation of the line between the points (-2,4) and (1,-2)

• Let \vec a be the position vector of A from the origin:
  • \vec A = \begin{pmatrix}-2 \\ 4\end{pmatrix}
• \vec{AB} = \vec B - \vec A
  • = \begin{pmatrix}1 \\ -2\end{pmatrix} - \begin{pmatrix}-2 \\ 4\end{pmatrix}
  • = \begin{pmatrix}3 \\ -6\end{pmatrix}
• So the direction vector is \begin{pmatrix}3 \\ -6\end{pmatrix}, which we can simplify to \begin{pmatrix}1 \\ -2\end{pmatrix} (because it’s just a direction and the magnitude is not important here).
• Equation of a line:
  • \vec r = \vec a + \lambda \vec d
  • = \begin{pmatrix} -2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \end{pmatrix}
• Answer: \vec r = \begin{pmatrix} -2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \end{pmatrix}

More complex exam-style questions

A is the point (3,1,4) and AQ=18. Find Q if A and Q are on the line r=\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} + \lambda \begin{pmatrix}4 \\ -2 \\ -4\end{pmatrix}

• Find the length of vector \begin{pmatrix}4 \\ -2 \\ -4\end{pmatrix}:
  • |\vec{d}| = \sqrt{4^2 + (-2)^2 + (-4)^2} = \pm\sqrt{16 + 4 + 16} = \pm\sqrt{36} = \pm6
  • \lambda\times|\begin{pmatrix}4 \\ -2 \\ -4\end{pmatrix}|=18
  • \lambda \times \pm6 = 18
  • \lambda = \pm3
• Find Q:
  • Q=\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} \pm 3 \begin{pmatrix}4 \\ -2 \\ -4\end{pmatrix}
  • =\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} \pm \begin{pmatrix}12 \\ -6 \\ -12\end{pmatrix}
  • =\begin{pmatrix}13 \\ -4 \\ -9\end{pmatrix} or \begin{pmatrix}-11 \\ 8 \\ 15\end{pmatrix}
• Answer: Q=\begin{pmatrix}13 \\ -4 \\ -9\end{pmatrix} or Q=\begin{pmatrix}-11 \\ 8 \\ 15\end{pmatrix}

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