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Section 11.2 : Vector Arithmetic

1. Given \(\vec a = \left\langle {8,5} \right\rangle \) and \(\vec b = \left\langle { - 3,6} \right\rangle \) compute each of the following.
  1. \(6\vec a\)
  2. \(7\vec b - 2\vec a\)
  3. \(\left\| {10\vec a + 3\vec b} \right\|\)
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a \(6\vec a\) Show Solution

This is just a scalar multiplication problem. Just remember to multiply each component by the scalar, 6 in this case.

\[6\vec a = 6\left\langle {8,5} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {48,30} \right\rangle }}\]

b \(7\vec b - 2\vec a\) Show Solution

Here we’ll just do each of the scalar multiplications and then do the subtraction. With the subtraction just remember to subtract corresponding components from each vector and recall that order is important here since we are doing subtraction!

\[7\vec b - 2\vec a = 7\left\langle { - 3,6} \right\rangle - 2\left\langle {8,5} \right\rangle = \left\langle { - 21,42} \right\rangle - \left\langle {16,10} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle { - 37,32} \right\rangle }}\]

c \(\left\| {10\vec a + 3\vec b} \right\|\) Show Solution

So, first we compute the vector inside the magnitude bars and the compute the magnitude.

\[10\vec a + 3\vec b = 10\left\langle {8,5} \right\rangle + 3\left\langle { - 3,6} \right\rangle = \left\langle {80,50} \right\rangle + \left\langle { - 9,18} \right\rangle = \left\langle {71,68} \right\rangle \]

The magnitude is then,

\[\left\| {10\vec a + 3\vec b} \right\| = \sqrt {{{\left( {71} \right)}^2} + {{\left( {68} \right)}^2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt {9665} }}\]