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Scalar Product of Vectors Theory
![Given \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}=x_1 i+y_1 j\) and \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}=x_2 i+y_2 j\) then \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} \cdot \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}=x_1 x_2+y_1 y_2\) and \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} \cdot \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}=|a||b| \cos \theta\). If \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}\) is perpendicular to \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}\) then \(\theta=\dfrac{\pi}{2}\).\\ \(a \cdot b=|a||b| \cos \dfrac{\pi}{2}\).\\ \(\therefore a \cdot b=0\)\\ \begin{multicols}{2} \textbf{Example 1}\\ %question 12287 The scalar product \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} \cdot \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b} \) of \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} = 3\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{i} + 2\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{j} \) and \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b} = - 2\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{i} + 5\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{j} \) is?\\ \textbf{Example 1 solution}\\ $\begin{aligned}[t] a \cdot b &=(3 i+2 j)(-2 i+5 j) \\ &=3 \times -2+2 \times 5 \\ &=4 \end{aligned}$\\ \columnbreak \textbf{Example 2}\\ The angle between \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} = 2\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{i} + 3\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{j} \) and \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b} = \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{i} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{j} \) is closest to?\\ \textbf{Example 2 solution}\\ $\begin{aligned}[t] &\begin{aligned} \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} \cdot \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b} &=(2i+3j)(i-j) \\ &=2-3 \\ &=-1\\ |\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}| &=\sqrt{4+9} \\ &=\sqrt{13}\\ |\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}| &=\sqrt{1+1} \\ &=\sqrt{2}\\ \end{aligned}\\ &\begin{aligned} \cos \theta&=\frac{a \cdot b}{|a|\times|b|}\\ &=\frac{-1}{\sqrt{26}}\\ \theta&=101^{\circ} \end{aligned} \end{aligned}$\\ \end{multicols}](/media/ytqahrkn/scalar-product-of-vectors.png)
NSW Y12 Maths - Extension 1 Vectors Scalar Product of Vectors
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