4.4 The Scalar or Dot Product of Two Vectors
The dot or scalar product of two non zero vectors A and B is the number \(\textbf {A}\cdot \textbf {B} = \left |\textbf {A}\right |\left |\textbf {B}\right |\hspace {0.1cm}\cos \theta \hspace {0.2cm},\)
\( \hspace {0.2cm} 0 \leq \theta \leq \pi ,\) where \(\theta \) is the angle between \(\textbf {A}\) and \(\textbf {B}\). If either \(\textbf {A}\) or \(\textbf {B}\) is 0, we define \(\textbf {A}\cdot \textbf {B} = 0\).
Work and the Dot Product
Let F be a force which moves an object from a point \(P\) to \(Q\), then the work done by F in moving the object is defined by \[W = \left |\overrightarrow {PQ}\right |\left |\textbf {F}\right | \cos \theta \]
Note that \(W = \overrightarrow {PQ}\cdot \textbf {F}\)
- 1.
- If the vectors \(\overline {a}\) and \(\overline {b}\) have length 4 and 6 angle between them is \(\dfrac {\pi }{3}\), find \(\overline {a}\cdot \overline {b}.\)
- 2.
- A crate is hauled 8\(m\) up a ramp under a constant force of \(200 N\) applied on an angle of \(25^o\) to the ramp. Find the work done.
Solution. \begin {align*} \overline {a}\cdot \overline {b} & = \left |\overline {a}\right |\left |b\right |\cos \theta \\ & = 4\cdot 6\cdot \cos \dfrac {\pi }{3}\\ & = 12 \end {align*}
\begin {align*} W & = \left |\textbf {F}\right |\left |\textbf {D}\right |\cos \theta \\ & = 200 \times 8 \times \cos 25^o\\ & = 1450 N\cdot M = 1450 J\\ \end {align*}
The Dot Product in Component form
The dot product of \(\overline {a} = \langle a_1, a_2, a_3\rangle \) and \(\overline {b} = \langle b_1, b_2, b_3\rangle \) is \[\overline {a}\cdot \overline {b} = a_1b_1 + a_2b_2 + a_3b_3\]
- 1.
- Show that \(2\textbf {i} + 2\textbf {j} -\textbf {k}\) is perpendicular to \( 5\textbf {i} - 4\textbf {j} + 2\textbf {k}\).
- 2.
- Find the angle between the vectors \(\overline {a} = \langle 2,2,-1\rangle \) and \(\overline {b} = \langle 5, -3,2\rangle \).
- 3.
- A force is given by a vector \(\textbf {F} = 3\textbf {i} + 4\textbf {j} + 5\textbf {k}\) and moves a particle from the point \(P(2,1,0)\) to \(Q(4,6,2)\). Find the work done.
Solution.
Part 1
Two vectors are perpendicular if their dot product is 0 \[(2\textbf {i} + 2\textbf {j} -\textbf {k})\cdot ( 5\textbf {i} - 4\textbf {j} + 2\textbf {k}) = 10 - 8 - 2 = 0\] \(\therefore \) the two vectors are perpendicular.
Part 2
\(\displaystyle {\overline {a}\cdot \overline {b} = \left |\overline {a}\right |\left |\overline {b}\right |\cos \theta \implies \cos \theta = \frac {\overline {a}\cdot \overline {b}}{ \left |\overline {a}\right |\left |\overline {b}\right | }}\)
\(\displaystyle {\overline {a}\cdot \overline {b} = 2(5) - 2(3) -1(2) = 2}\)
\(\displaystyle {\left |\overline {a}\right | = \sqrt {9} = 3\hspace {1cm} , \hspace {1cm} \left |\overline {b}\right |= \sqrt {38}}\)
\begin {align*} \cos \theta & = \frac {2}{3\sqrt {38}}\implies \theta = \cos ^{-1} \Big (\frac {2}{3\sqrt {38}}\Big ) \approx 84^o\\\\ \end {align*}
Part 3
\begin {align*} W & = \textbf {F}\cdot (\overrightarrow {PQ})\\ & = (\textbf {F} = 3\textbf {i} + 4\textbf {j} + 5\textbf {k})\cdot ( 2\textbf {i} + 5\textbf {j} + 2\textbf {k})\\\\ & = 36\\ \end {align*}
Properties of Dot Product
If \(\overline {a}, \overline {b}\) and \(\overline {c}\) are vectors and \(\alpha \) is a scalar, then
- 1.
- \(\overline {a}\cdot \overline {a} = \left |\overline {a}\right |^2\)
- 2.
- \(\overline {a}\cdot \overline {b} = \overline {b}\cdot \overline {a}\)
- 3.
- \(\overline {a}\cdot (\overline {b}+ \overline {c}) = \overline {a}\cdot \overline {b} + \overline {a}\cdot \overline {c}\)
- 4.
- \((\alpha \overline {a})\cdot \overline {b} = \alpha (\overline {a}\cdot \overline {b}) = \overline {a}\cdot (\alpha \overline {b})\)
- 5.
- \(\overline {0}\cdot \overline {a} = \overline {a}\cdot \overline {0} = 0\)
Proof. □
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