Resources for Vectors in Geometry
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Questions
11
With Worked SolutionClick Here -
Video Tutorials
2
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HSC Questions
4
With Worked SolutionClick Here
Vectors in Geometry Theory
![Many geometry theorems can be solved using vector methods.\\ \textbf{Example}\\ \begin{center} \resizebox{0.25\textwidth}{!}{ \begin{tikzpicture} \coordinate [label=above:$A$ ] (A) at (-2,4); \coordinate [ label=below:$B$ ] (B) at (2,0); \coordinate [ label=below:$C$ ] (C) at (-2,0); \draw[fill=black] (-2,2) node[left=1mm]{$\underset{\sim}b$}; \draw[fill=black] (0,0) node[below=1mm]{$\underset{\sim}a$}; \draw[fill=black] (0,2) node[right=1mm,yshift=1mm]{$\underset{\sim}c$}; \draw[line width=1pt] (A)--(B)--(C)--cycle; \draw[decoration={markings, mark=at position 0.7 with {\arrow[scale=1.3]>}}, postaction=decorate,line width=1pt](A)--(B); \draw[decoration={markings, mark=at position 0.4 with {\arrow[scale=1.3]<}}, postaction=decorate,line width=1pt](B)--(C); \draw[decoration={markings, mark=at position 0.3 with {\arrow[scale=1.3]<}}, postaction=decorate,line width=1pt](C)--(A); \draw[line width=1pt] (C) +(.25,0) |- +(0,.25); \end{tikzpicture} } %\includegraphics[width=0.2\textwidth]{92ea7b61-8fc8-4d55-9af3-966e60398ba3} \end{center} In \(\triangle A B C\) given that \(|\underset{\sim}c|^2=|\underset{\sim}a|^2+|\underset{\sim}b|^2\), prove that \(\angle A C B=\dfrac{\pi}{2}\)\\ \textbf{Solution}\\ $\begin{aligned} {\underset{\sim}c}&=\underset{\sim}b+\underset{\sim}a \\ |\underset{\sim}c|^2&=c \cdot c=(b+a)(b+a) \\ &= b \cdot b+2 a \cdot b+a \cdot a \\ |c|^2&=|b|^2+|a|^2+2 a \cdot b \\ |a|^2+|b|^2&=|b|^2+|a|^2+2 a \cdot b \\ \therefore 2 a \cdot b&=0 \\ \therefore a \cdot b&=0 \\ \therefore \angle A C B&=\frac{\pi}{2} \end{aligned}$\\](/media/awnpe4lo/vectors-in-geometry.png)