Vectors in cartesian form

When vectors are given in cartesian form there is an alternative formula for calculating the scalar product.

Proof. Consider the vector $b - a = (\begin{matrix} b_{1} - a_{1} \\ b_{2} - a_{2} \end{matrix})$ . The modulus of this is

| b - a | = \sqrt{{(b_{1} - a_{2})}^{2} + {(b_{2} - a_{2})}^{2}} .

Note from figure 2 that the vectors $a$ , $b$ and $b - a$ form a triangle:

Two vectors, labelled a and b, are drawn as line segments. Their start points coincide and their end points do not. There is an acute angle of theta between them. Forming a triangle, a third line segment, labelled b-a is added. This vector starts at the end point of a and ends at the end point of b. Each line segment has an arrow on it pointing from the start point to the end point.

Figure 2: A triangle is formed by two vectors and their difference.

Let $θ$ denote the angle between $a$ and $b$ . Then, the cosine rule yields:

{| b - a |}^{2} = {| a |}^{2} + {| b |}^{2} - 2 | a | | b | cos θ .

(1)

Substituting the definition of the scalar product of $a$ and $b$ into equation 1 gives:

{| b - a |}^{2} = {| a |}^{2} + {| b |}^{2} - 2 (a \cdot b) .

Rearranging:

(a \cdot b) = \frac{1}{2} ({| a |}^{2} + {| b |}^{2} - {| b - a |}^{2}) .

Writing this in terms of components produces:

\begin{array}{l} (a \cdot b) & = \frac{1}{2} (a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} - {(b_{1} - a_{1})}^{2} - {(b_{2} - a_{2})}^{2}) \\ = \frac{1}{2} (a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} - b_{1}^{2} + 2 b_{1} a_{1} - a_{1}^{2} - b_{2}^{2} + 2 b_{2} a_{2} - a_{2}^{2}) \\ = \frac{1}{2} (2 b_{1} a_{1} + 2 b_{2} a_{2}) \\ = a_{1} b_{1} + a_{2} b_{2} \end{array}

as required. □

Example 1.5.

Consider again the vectors

a = (\begin{matrix} 2 \\ 2 \end{matrix}) and b = (\begin{matrix} 4 \\ 0 \end{matrix}) .

Calculating the scalar product using the components:

a \cdot b = a_{1} b_{1} + a_{2} b_{2} = 2 \times 4 + 2 \times 0 = 8.

Note that if we are given vectors in this form, the scalar product may be used to calculate the angle between them. Since $a \cdot b = 8$ and we have:

\begin{array}{l} | a | & = \sqrt{8} \\ | b | & = 4 . \end{array}

Hence,

\begin{array}{l} 8 = a \cdot b & = | a | | b | cos θ \\ = 4 \sqrt{8} cos θ . \end{array}

Rearranging:

θ = {cos}^{- 1} (\frac{8}{4 \sqrt{8}}) = 45^{\circ} .

1.1 Vectors in cartesian form