Section 11.1 : Basic Concepts
3. Give the vector for the position vector for \(\left( { - 3,2,10} \right)\). Find its magnitude and determine if the vector is a unit vector.
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Start SolutionWriting down a vector for a line segment is really simple. Just recall that the components of the vector are always the coordinates of the ending point minus the coordinates of the starting point.
Just recall that the starting point for any position vector is the origin and the ending point is the point we’re working with. In other words, the components of the position vector are simply the coordinates of the point.
Here is the position vector for this point.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec v = \left\langle { - 3,2,10} \right\rangle }}\] Show Step 2To compute the magnitude just recall the formula we gave in the notes. The magnitude of this vector is then,
\[\left\| {\vec v} \right\| = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( {10} \right)}^2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt {113} }}\] Show Step 3Because we can see that \(\left\| {\vec v} \right\| = \sqrt {113} \ne 1\) we know that this vector is not a unit vector.