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Properties of the Scalar Product0%

Vectors · Topic 11 of 11

Properties of the Scalar Product

Video lesson · from 53:442 worked examples

One lesson video covers all of Vectors, so it opens at 53:44 for this topic — not from the beginning.

Theory

1. The scalar product is commutative i.e. a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

2. The scalar product is distributive i.e. a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}

3. The scalar product of a vector and itself is a positive real number if a⃗≠0\vec{a} \neq 0 i.e. a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2.

Proof: a⃗⋅a⃗=∣a⃗∣∣a⃗∣cos⁡θ\vec{a} \cdot \vec{a} = |\vec{a}| |\vec{a}| \cos \theta. Since the angle is 0, cos⁡0∘=1\cos 0^\circ = 1. a⃗⋅a⃗=∣a⃗∣∣a⃗∣×1=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}| |\vec{a}| \times 1 = |\vec{a}|^2.

⚠️ Common Examiner Traps

  • a⃗⋅a⃗=∣a⃗∣2\vec{a}\cdot\vec{a} = |\vec{a}|^2: the scalar product of a vector with itself is the square of its magnitude, which is often the quickest route into a proof.
  • It distributes over addition: a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}, which lets you expand exactly like ordinary algebra.
  • Order does not matter: a⃗⋅b⃗=b⃗⋅a⃗\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}.
  • You cannot divide by a vector: there is no such operation, so rearranging a scalar product equation must be done by expanding, not dividing.

Worked examples

Example 1

Calculate p⃗⋅(q⃗+r⃗)\vec{p} \cdot (\vec{q} + \vec{r}) when ∣p⃗∣=3|\vec{p}| = 3, ∣q⃗∣=4|\vec{q}| = 4 and ∣r⃗∣=2|\vec{r}| = 2, with the angle between p⃗\vec{p} and q⃗\vec{q} being 60∘60^\circ and the angle between p⃗\vec{p} and r⃗\vec{r} being 120∘120^\circ.

p⃗⋅(q⃗+r⃗)=p⃗⋅q⃗+p⃗⋅r⃗=∣p⃗∣∣q⃗∣cos⁡60∘+∣p⃗∣∣r⃗∣cos⁡120∘=(3)(4)(0.5)+(3)(2)(−0.5)=6−3=3\begin{aligned} \vec{p} \cdot (\vec{q} + \vec{r}) &= \vec{p} \cdot \vec{q} + \vec{p} \cdot \vec{r} \\ &= |\vec{p}||\vec{q}|\cos 60^\circ + |\vec{p}||\vec{r}|\cos 120^\circ \\ &= (3)(4)(0.5) + (3)(2)(-0.5) \\ &= 6 - 3 \\ &= 3 \end{aligned}

Example 2

If ∣a⃗∣=∣c⃗∣=2|\vec{a}| = |\vec{c}| = 2 and ∣b⃗∣=23|\vec{b}| = 2\sqrt{3}, angle between a⃗\vec{a} and b⃗\vec{b} is 30∘30^\circ, angle between a⃗\vec{a} and c⃗\vec{c} is 150∘150^\circ. Calculate a⃗⋅(a⃗+b⃗+c⃗)\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}).

a⃗⋅(a⃗+b⃗+c⃗)=a⃗⋅a⃗+a⃗⋅b⃗+a⃗⋅c⃗=∣a⃗∣2+∣a⃗∣∣b⃗∣cos⁡30∘+∣a⃗∣∣c⃗∣cos⁡150∘=22+(2)(23)(32)+(2)(2)(−32)=4+6−23=10−23\begin{aligned} \vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}) &= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \\ &= |\vec{a}|^2 + |\vec{a}||\vec{b}|\cos 30^\circ + |\vec{a}||\vec{c}|\cos 150^\circ \\ &= 2^2 + (2)(2\sqrt{3})\left(\frac{\sqrt{3}}{2}\right) + (2)(2)\left(-\frac{\sqrt{3}}{2}\right) \\ &= 4 + 6 - 2\sqrt{3} \\ &= 10 - 2\sqrt{3} \end{aligned}