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 \(\theta\) is the angle between the two vectors is defined by:
\(\vec{a}\cdot\vec{b} = \vert \vec{a} \vert \vert \vec{b} \vert \cos{θ}\)
\(\vec{u}\cdot\vec{u}=0 \) if \(\vec{u}\perp\vec{v} \)
\(\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u} \)
\(\vec{u}\cdot\vec{u}=\vert\vec{u}\vert^2 \)
\(\vec{u}\cdot\vec{0}=0 \)
\(\vec{u}\cdot\left(\vec{a}+\vec{b}\right)=\vec{u}\cdot\vec{a}+\vec{u}\cdot\vec{b} \)
\(k\left(\vec{u}\cdot\vec{v}\right)=\left(\vec{ku}\right)\cdot\vec{v}=\vec{u}\cdot\left(\vec{ku}\right) \)
\(\vert\vec{u}\vert=7\), \(\vert\vec{v}\vert=12\), \(θ=60°\)
We can determine the dot product of this vector pair as such:
\(\vec{u}\cdot\vec{v} = \vert \vec{u} \vert \vert \vec{v} \vert \cos{θ}\)
\(\vec{u}\cdot\vec{v}=(7)(12)\cos{(60°)}\)
\(\vec{u}\cdot\vec{v} = 42\)
Therefore, we can determine the dot product is \(\boldsymbol{42}\).
\(\vert\vec{a}\vert=20\), \(\vert\vec{b}\vert=3\), \(θ=\dfrac{5\pi}{6}\)
We can determine the dot product of this vector pair as such:
\(\vec{a}\cdot\vec{b} = \vert \vec{a} \vert \vert \vec{b} \vert \cos{θ}\)
\(\vec{a}\cdot\vec{b}=(20)(3)\cos{\left(\dfrac{5\pi}{6}\right)}\)
\(\vec{a}\cdot\vec{b} = 60\times\left(-\dfrac{\sqrt{3}}{2}\right)\)
\(\vec{a}\cdot\vec{b} = -30\sqrt{3}\)
Therefore, we can determine the dot product is \(\boldsymbol{-30\sqrt{3}}\).
\(\vec{v}\cdot\vec{u}=(12)(7)\cos{(60°)} = 42\)
\(\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}\)
Therefore, we can conclude that dot product is commutative.
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, we can conclude that \(\boldsymbol{\vec{a}\cdot\vec{a}=\vert\vec{a}\vert^2 }\).
We can assess these dot products as such:
\(\vec{a}\cdot\vec{0}=(7)(0)\cos{\theta} = 0\)
\(\vec{v}\cdot\vec{0}=(12)(0)\cos{\theta} = 0\)
Therefore, we can determine that the dot product of zero vector and any vector is \(\boldsymbol{0}\).
Proof 1:
Assume \(\vec{u}\perp\vec{v} \; \text{then} \; \theta = 90°\)
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:
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:
\(\cos{\theta} = 0\)
\(\theta = 90°\)
Therefore, we can determine \(\vec{u} \) and \(\vec{v} \) have an angle of \(90°\) between them.
\(\left(\vec{a}\cdot\vec{b}\right)\cdot\vec{c}\neq\vec{a}\cdot\left(\vec{c}\cdot\vec{b}\right)\)
\(\left(\vec{a}\cdot\vec{b}\right)\) is a scalar. Dot product requires two vectors not a vector and a scalar as shown above.