VECTORS

In three dimensions, the base vectors are
\(i=\begin{pmatrix} 1\\0 \\0 \end{pmatrix}\), \(j=\begin{pmatrix} 0\\1 \\0 \end{pmatrix}\), \(k=\begin{pmatrix} 0\\0 \\1 \end{pmatrix}\)
that are along the \(x, y\) and \(z\) coordinate directions, respectively, as shown in the figure.
For example, \(v=\begin{pmatrix} -4\\3 \\2 \end{pmatrix}\) can be written as \(-4i+3j+2k\).
The magnitude of \(v\):

\(|v|=\sqrt{((-4)^2+3^2+2^2)}=\sqrt{29}\)
Given that if \(v=xi+yj+zk\) then
\(|v|=\sqrt{x^2+y^2+z^2}\)

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UNIT VECTORS

If \(\overrightarrow{AB}\) is a non-unit vector, then a unit vector in the direction of \(\overrightarrow{AB}\) is

\({1\over |\overrightarrow{AB}|}\overrightarrow{AB}\)
LESSON 1
Determine the unit vector in the direction of the vector \(\overrightarrow{OA}=\begin{pmatrix} -4\\3 \\2 \end{pmatrix}\).
SOLUTION
  • Determine the length of the vector
\(|\overrightarrow{OA}|=\sqrt{(-4)^2+(3)^2+(2)^2}=\sqrt{29}\)
  • Use the form for a unit vector
Unit Vector
\({1\over \sqrt{29}} \begin{pmatrix} -4\\3 \\2 \end{pmatrix}\)

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Complete Question 1 from worksheet.