AHL 3.13—Scalar (dot) product

The notation uses a bold dot (·) between the two vectors. Never write $\mathbf{a}\times \mathbf{b}$ for the scalar product , that is reserved for the vector (cross) product, which gives a vector, not a scalar.

Example 1: Let $\mathbf{a}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right)$. Find $\mathbf{a}\cdot \mathbf{b}$.

Step 1: Multiply corresponding components.
$\mathbf{a}\cdot \mathbf{b}=(1)(4)+(2)(5)+(3)(6)$

Step 2: Add the products.
$=4+10+18=32$

Note that $\mathbf{b}\cdot \mathbf{a}=(4)(1)+(5)(2)+(6)(3)=4+10+18=32$ , the same result, illustrating commutativity.

Example 2: Let $\mathbf{u}=\left(\begin{array}{c}-3\\ 0\\ 2\end{array}\right)$ and $\mathbf{v}=\left(\begin{array}{c}1\\ 5\\ 4\end{array}\right)$. Find $\mathbf{u}\cdot \mathbf{v}$.

$\mathbf{u}\cdot \mathbf{v}=(-3)(1)+(0)(5)+(2)(4)=-3+0+8=5$

Notice that the middle term vanishes immediately because $0\times 5=0$. Always check for zeros to speed up your calculation.

Work left to right: multiply each pair, write down the three products, then sum them. Do not try to do it all in one mental step , sign errors are very common under exam pressure.