Vector multiplication can be completed using the dot product or cross product.
The result of the dot product is a scalar. We can calculate the dot product using geometric or algebraic vectors.
The geometric dot product of two vectors \(\vec{a}\) and \(\vec{b}\) where θ is the angle between the two vectors
is defined by:
\( \vec{a}\cdot\vec{b} = \vert \vec{a} \vert \vert \vec{b} \vert \cos{θ} \)
Dot Product Properties
\(\vec{u}\cdot\vec{u}=0 \) iff \(\vec{u}\perp\vec{v} \)
\(\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u} \)
\(\vec{u}\cdot\vec{u}=\vert\vec{u}\vert^2 \)
\(\vec{u}\cdot\vec{0}=0 \)
\(\vec{u}\cdot\left(\vec{a}+\vec{b}\right)=\vec{u}\cdot\vec{a}+\vec{u}\cdot\vec{b} \)
\(k\left(\vec{u}\cdot\vec{v}\right)=\left(\vec{ku}\right)\cdot\vec{v}=\vec{u}\cdot\left(\vec{ku}\right)
\)
Find the dot product for \(\vec{u}\cdot\vec{v} \) for each of the following where θ is the
angle between vectors.
\(\vert\vec{u}\vert=7\), \(\vert\vec{v}\vert=12\), \(θ=60° \)
\(\vert\vec{a}\vert=20\), \(\vert\vec{b}\vert=3\), \(θ=\dfrac{5\pi}{6} \)
Show Answer
\(\vec{u}\cdot\vec{v}=(7)(12)\cos{(60°)}=42 \)
\( \vec{a}\cdot\vec{b}=(20)(3)\cos{(\dfrac{5\pi}{6})} = 60\times\left(-\dfrac{\sqrt{3}}{2}\right) = -30\sqrt{3} \)
For the above question find \(\vec{v}\cdot\vec{u} \). What is property
you can conclude from this?
Show Answer
\(\vec{v}\cdot\vec{u}=(12)(7)\cos{(60°)}=42 \)
∴\(\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u} \)
ie. dot product is commutative.
Find \(\vec{a}\cdot\vec{a} \) and \(\vec{b}\cdot\vec{b} \). What can you conclude from this?
Show Answer
Since the angle between \(\vec{a} \) and itself is 0°:
\(\vec{a}\cdot\vec{a}= \vert \vec{a} \vert \vert \vec{a} \vert \cos{(0°)}={\vert \vec{a} \vert}^2 \)
\(\vec{b}\cdot\vec{b}=\vert \vec{b} \vert \vert \vec{b} \vert \cos{(0°)} = {\vert \vec{b} \vert}^2 = 3^2=9 \)
Therefore, \(\vec{a}\cdot\vec{a}=\vert\vec{a}\vert^2 \).
Find \(\vec{a}\cdot\vec{0} \) and \(\vec{v}\cdot\vec{0} \). What can conclude from this?
Show Answer
\(\vec{u}\cdot\vec{0}=(7)(0)\cos{θ}=0 \)
\(\vec{u}\cdot\vec{0}=(12)(0)\cos{θ}=0 \)
Dot product of zero vector and any vector is 0.
Prove that two non-zero vectors \(\vec{u} \) and \(\vec{v} \) are perpendicular, if and only
if \(\vec{u}\cdot\vec{v}=0 \).
Proof 1:
Show Answer
assume \(\vec{u}\perp\vec{v}\text{ then θ=90° } \)
then \(\vec{u}\cdot\vec{v} = \vert\vec{u}\vert\vert\vec{v}\vert\cos{90°} \)
\(\vert\vec{u}\vert\vert\vec{v}\vert(0) = 0\)
Proof 2:
Show Answer
assume \(\vec{u}\cdot\vec{v}=0 \)
then \(\vert\vec{u}\vert\vert\vec{v}\vert\cos{θ}=0 \)
Since the vectors are non-zero,
\(\cos{θ}=0 \)
\(θ=90° \)
ie. \(\vec{u} \) and \(\vec{v} \) have an angle of 90° between them.
Explain why
\(\left(\vec{a}\cdot\vec{b}\right)\cdot\vec{c}\neq\vec{a}\cdot\left(\vec{c}\cdot\vec{b}\right)
\)
Show Answer
\(\left(\vec{a}\cdot\vec{b}\right)\cdot\vec{c}\neq\vec{a}\cdot\left(\vec{c}\cdot\vec{b}\right)
\)
\(\left(\vec{a}\cdot\vec{b}\right) \) is a scalar. Dot product requires two vectors not a vector and a scalar as shown above.