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Question

If

AB=(132),AC=(214) and BD=(023)\overrightarrow{\mathrm{AB}}=\left( \begin{array}{c}{-1} \\ {3} \\ {2}\end{array}\right), \overrightarrow{\mathrm{AC}}=\left( \begin{array}{c}{2} \\ {-1} \\ {4}\end{array}\right) \quad \text { and } \quad \overrightarrow{\mathrm{BD}}=\left( \begin{array}{c}{0} \\ {2} \\ {-3}\end{array}\right)

find: a. AD\overrightarrow{\mathrm{AD}} b. CB\overrightarrow{\mathrm{CB}} c. CD\overrightarrow{\mathrm{CD}}

Solution

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AB=(132),AC=(214),BD=(023),(Given)\overrightarrow {AB}=\begin{pmatrix}-1\\3\\2\end{pmatrix}, \qquad \overrightarrow {AC}=\begin{pmatrix}2\\-1\\4\end{pmatrix}, \qquad \overrightarrow {BD}=\begin{pmatrix}0\\2\\-3\end{pmatrix}, \qquad \qquad (Given)

(a)\textbf{(a)}

AD=AB+BDAlternative path from A to D=(132)+(023)Add vectorsAD=(151)\begin{align*} \overrightarrow {AD} & =\overrightarrow {AB}+\overrightarrow {BD} && \text {Alternative path from $A$ to $D$} \\ & =\begin{pmatrix}-1\\3\\2\end{pmatrix}+\begin{pmatrix}0\\2\\-3\end{pmatrix} && \text {Add vectors} \\ \overrightarrow {AD} & =\begin{pmatrix}-1\\5\\-1\end{pmatrix} \\ \end{align*}

(b)\textbf{(b)}

CB=CA+ABAlternative path from C to BCA=ACNegative vectorCB=(214)+(132)Add vectorsCB=(342)\begin{align*} \overrightarrow {CB} & =\overrightarrow {CA}+\overrightarrow {AB} && \text {Alternative path from $C$ to $B$} \\ \overrightarrow {CA} & =-\overrightarrow {AC} && \text {Negative vector} \\ \overrightarrow {CB} & =-\begin{pmatrix}2\\-1\\4\end{pmatrix}+\begin{pmatrix}-1\\3\\2\end{pmatrix} && \text {Add vectors} \\ \overrightarrow {CB} & =\begin{pmatrix}-3\\4\\-2\end{pmatrix} \\ \end{align*}

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