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NSW Y12 Maths Standard 1 NSW Y12 Maths Standard 2 NSW Y12 Maths Advanced NSW Y12 Maths Extension 1 NSW Y12 Maths Extension 2

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Complex Numbers Proof Further Work With Vectors Integration Mechanics

NSW Y12 Maths - Extension 2 Further Work With Vectors Applications of the Dot Product

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Applications of the Dot Product Theory

Applications of the Dot Product - The scalar projection of \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}\) onto \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}\) is \(a \cdot \hat{b}\) where \(a \cdot \hat{b}=\dfrac{a \cdot b}{|b|}\)\\ The vector projection of \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}\) onto \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}\) is \((a \cdot \hat{b}) \hat{b}\) or \(\dfrac{a \cdot b}{|b|} \times \hat{b}\)\\ The vector projection of \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}\) perpendicular to \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}\) is \(a-(a \cdot \hat{b}) \hat{b}\)\\ \begin{multicols}{2} \textbf{Example 1}\\ Given that \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}=3 i+2 j-k\) and \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}=2 i-j+k\) find:\\ i) the scalar projection of \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}\) onto \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}\).\\ ii) the vector projection of \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}\) outer \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}\).\\ iii) the vector projection of \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a}\) perpendicular to \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b}\).\\ \textbf{Example 1 solution}\\ $\begin{aligned} \text { i) } \text { Scalar projection } & =\frac{a \cdot b}{|b|} \\ & =\frac{6-2-1}{\sqrt{4+1+1}} \\ & =\frac{3}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} \\ & =\frac{\sqrt{6}}{2} \end{aligned}$\\ $\begin{aligned} \text { ii) } \text { Vector projection } & =\frac{a \cdot b}{\mid b |} \times \hat{b} \\ & =\frac{\sqrt{6}}{2} \times \frac{(2 i-j+k)}{\sqrt{6}} \\ & =\frac{1}{2}(2 i-j+k) \end{aligned}$\\ $\begin{aligned} \text { iii) vector}&\text{ projection of } \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} \perp \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{b} \\ & =3 i+2 j-k-\frac{1}{2}(2 i-j+k) \\ & =2 i+\frac{5}{2} j-\frac{3}{2} k \end{aligned}$\\ \columnbreak \textbf{Example 2}\\ Given \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{u}=i+j+2 k\) and \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{v}=2 i-j+k\) find unit vectors perpendicular to \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{u}\) and \(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{v}\).\\ \textbf{Example 2 solution}\\ $\begin{aligned} & \text { Let } \omega=p i+q j+r k \text { be the unit vector. } \\ & u \cdot w=0 \qquad p+q+2 r=0 \ldots (1) \\ & \qquad\qquad \qquad 2 p-q+r=0 \ldots (2) \\ & \qquad \qquad \qquad p^2+q^2+r^2=1 \ldots \text { (3) } \\ & (1)+(2) \qquad 3 p+3 r=0 \\ & \qquad \qquad \qquad \qquad p=-r \\ & \text { In (1) }\qquad -r+q+2 r=0 \\ & \qquad \qquad \qquad \qquad \qquad q=-r \\ & \text { In (3) } \qquad r^2+r^2+r^2=1 \\ & \qquad \qquad \qquad \qquad 3 r^2=1 \\ & \qquad \qquad \qquad \qquad r=\pm \frac{1}{\sqrt{3}} \\ & p=\mp \frac{1}{\sqrt{3}}, q=\mp \frac{1}{\sqrt{3}}, r=\pm \frac{1}{\sqrt{3}} \\ & \qquad \therefore \omega=\mp \frac{1}{\sqrt{3}}(i+j-k) \end{aligned}$\\ \end{multicols}

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