A formula for finding the scalar product

2 A formula for finding the scalar product

We can use the results summarized in Key Point 10 to obtain a formula for finding a scalar product when the vectors are given in Cartesian form. We consider vectors in the $x y$ plane. Suppose $\underset{̲}{a} = a_{1} \underset{̲}{i} + a_{2} \underset{̲}{j}$ and $\underset{̲}{b} = b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j}$ . Then

\begin{array}{rcl} \underset{̲}{a} \cdot \underset{̲}{b} & = & (a_{1} \underset{̲}{i} + a_{2} \underset{̲}{j}) \cdot (b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j}) \\ = & a_{1} \underset{̲}{i} \cdot (b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j}) + a_{2} \underset{̲}{j} \cdot (b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j}) \\ = & a_{1} b_{1} \underset{̲}{i} \cdot \underset{̲}{i} + a_{1} b_{2} \underset{̲}{i} \cdot \underset{̲}{j} + a_{2} b_{1} \underset{̲}{j} \cdot \underset{̲}{i} + a_{2} b_{2} \underset{̲}{j} \cdot \underset{̲}{j} \end{array}

Using the results in Key Point 10 we can simplify this to give the following formula:

Key Point 12

If \underset{̲}{a} = a_{1} \underset{̲}{i} + a_{2} \underset{̲}{j} and \underset{̲}{b} = b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} then

\underset{̲}{a} \cdot \underset{̲}{b} = a_{1} b_{1} + a_{2} b_{2}

Thus to find the scalar product of two vectors their $\underset{̲}{i}$ components are multiplied together, their $\underset{̲}{j}$ components are multiplied together and the results are added.

Example 11

If $\underset{̲}{a} = 7 \underset{̲}{i} + 8 \underset{̲}{j}$ and $\underset{̲}{b} = 5 \underset{̲}{i} - 2 \underset{̲}{j}$ , find the scalar product $\underset{̲}{a} \cdot \underset{̲}{b}$ .

Solution

We use Key Point 12:

$\underset{̲}{a} \cdot \underset{̲}{b} = (7 \underset{̲}{i} + 8 \underset{̲}{j}) \cdot (5 \underset{̲}{i} - 2 \underset{̲}{j}) = (7) (5) + (8) (- 2) = 35 - 16 = 19$

The formula readily generalises to vectors in three dimensions as follows:

Key Point 13

If \underset{̲}{a} = a_{1} \underset{̲}{i} + a_{2} \underset{̲}{j} + a_{3} \underset{̲}{k} and \underset{̲}{b} = b_{1} \underset{̲}{i} + b_{2} \underset{̲}{j} + b_{3} \underset{̲}{k} then

\underset{̲}{a} \cdot \underset{̲}{b} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

Example 12

If $\underset{̲}{a} = 5 \underset{̲}{i} + 3 \underset{̲}{j} - 2 \underset{̲}{k}$ and $\underset{̲}{b} = 8 \underset{̲}{i} - 9 \underset{̲}{j} + 11 \underset{̲}{k}$ , find $\underset{̲}{a} \cdot \underset{̲}{b}$ .

Solution

We use the formula in Key Point 13:

$\underset{̲}{a} \cdot \underset{̲}{b} = (5) (8) + (3) (- 9) + (- 2) (11) = 40 - 27 - 22 = - 9$

Note again that the result is a scalar: there are no $\underset{̲}{i}$ ’s, $\underset{̲}{j}$ ’s, or $\underset{̲}{k}$ ’s in the answer.

Task!

If $\underset{̲}{p} = 4 \underset{̲}{i} - 3 \underset{̲}{j} + 7 \underset{̲}{k}$ and $\underset{̲}{q} = 6 \underset{̲}{i} - \underset{̲}{j} + 2 \underset{̲}{k}$ , find $\underset{̲}{p} \cdot \underset{̲}{q}$ .

Use Key Point 13:

Task!

If $\underset{̲}{r} = 3 \underset{̲}{i} + 2 \underset{̲}{j} + 9 \underset{̲}{k}$ find $\underset{̲}{r} \cdot \underset{̲}{r}$ . Show that this is the same as ${|\underset{̲}{r}|}^{2}$ .

$\underset{̲}{r} \cdot \underset{̲}{r} = (3 \underset{̲}{i} + 2 \underset{̲}{j} + 9 \underset{̲}{k}) \cdot (3 \underset{̲}{i} + 2 \underset{̲}{j} + 9 \underset{̲}{k}) = 3 \underset{̲}{i} \cdot 3 \underset{̲}{i} + 3 \underset{̲}{i} \cdot 2 \underset{̲}{j} + \dots = 9 + 0 + \dots = 94$ .

$|\underset{̲}{r}| = \sqrt{9 + 4 + 81} = \sqrt{94}$ , hence ${|\underset{̲}{r}|}^{2} = \underset{̲}{r} \cdot \underset{̲}{r}$ .

The above result is generally true:

Key Point 14

For any vector $\underset{̲}{r}$ , ${|\underset{̲}{r}|}^{2} = \underset{̲}{r} \cdot \underset{̲}{r}$

2 A formula for finding the scalar product

Answer

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