Chapter 1: Vectors, Lines and Planes

Section 1.3: Dot Product

Example 1.3.4

In the $\mathrm{xy}$-plane, obtain all vectors that are orthogonal to $\mathbf{A}\=3 \mathbf{i}\+2 \mathbf{j}$. Solution Mathematical Solution If $\mathbf{B}\=u \mathbf{i} \+v \mathbf{j}$ is orthogonal to $\mathbf{A}\=3 \mathbf{i}\+2 \mathbf{j}$, then $\mathbf{A}·\mathbf{B}\=3 u\+2 v\=0$ must hold. Solving this equation for, say, $v\=-3 u\/2$ gives   $\mathbf{B}\=\left[\begin{array}{c}u\\ v\end{array}\right]\=\left[\begin{array}{c}u\\ -3 u\/2\end{array}\right]\=u\left[\begin{array}{c}1\\ -3\/2\end{array}\right]\=2 \mathrm{\λ}\left[\begin{array}{c}1\\ -3\/2\end{array}\right]\=\mathrm{\λ}\left[\begin{array}{c}2\\ -3\end{array}\right]$  The vector B can be any multiple of $2 \mathbf{i}-3 \mathbf{j}$.   Maple Solution •  Enter A as per Table 1.1.1. •  Context Panel: Assign Name $\mathbf{A}\=⟨3\,2⟩$$\stackrel{\text{assign}}{\to }$ •  Enter B as per Table 1.1.1. •  Context Panel: Assign Name $\mathbf{B}\=⟨u\,v⟩$$\stackrel{\text{assign}}{\to }$ Determine $v\=v\left(u\right)$ from the equation $\mathbf{A}·\mathbf{B}\=0$  •  Common Symbols palette: Dot-product operator Set the dot product of A and B equal to zero.   •  Context Panel: Solve≻Isolate Expression for≻$v$   •  Context Panel: Substitute Into≻B   •  Context Panel: Evaluate at a Point≻u = 2*lambda $\mathbf{A}·\mathbf{B}\=0$ $3u+2v=0$ $\stackrel{\text{isolate for v}}{\to }$ $v=-\frac{3u}{2}$ $\stackrel{\text{substitute into}}{\to }$ $\left[\begin{array}{c}u\\ -\frac{3u}{2}\end{array}\right]$ $\stackrel{\text{evaluate at point}}{\to }$ $\left[\begin{array}{c}2\mathrm{\lambda }\\ -3\mathrm{\lambda }\end{array}\right]$

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