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202207051339 Solution to 1974-CE-AMATH-I-XX

(a) In the figure below, ABCDEF is a regular hexagon. Which one of the 6 vectors \overrightarrow{AD}, \overrightarrow{DA}, \overrightarrow{FC}, \overrightarrow{CF}, \overrightarrow{EB}, \overrightarrow{BE} is equal to \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DC}?

(b) \mathbf{u} and \mathbf{v} are unit vectors making an angle 60^\circ with each other as shown in the figure below. AB is a vector making an angle 30^\circ with \mathbf{u} and |\overrightarrow{AB}|=2. Express \overrightarrow{AB} in terms of \mathbf{u} and \mathbf{v}.

(c) In the figure below, \angle B=90^\circ and m(\overrightarrow{AB})=10. Calculate \overrightarrow{AB}\cdot\overrightarrow{AC}.

(a)

\begin{aligned} \mathbf{AD} & = -\mathbf{DA} \\ |\mathbf{AD}| & = |\mathbf{DA}| \\ \mathbf{FC} & = -\mathbf{CF} \\ |\mathbf{FC}| & = |\mathbf{CF}| \\ \mathbf{EB} & = -\mathbf{BE} \\ |\mathbf{EB}| & = |\mathbf{BE}| \\ \end{aligned}

Ans. \mathbf{FC} \textrm{\scriptsize{OR}} \overrightarrow{FC} by inspection.

Working.

\begin{aligned} &\quad \mathbf{AB} + \mathbf{BC} + \mathbf{DC} \\ & = (\mathbf{AB} + \mathbf{BC}) + \mathbf{DC} \\ & = \mathbf{AC} + \mathbf{DC} \\ \dots & \textrm{ as }\mathbf{AC}=\mathbf{FD} \enspace\dots \\ & = \mathbf{FD} + \mathbf{DC} \\ & = \mathbf{FC} \\ \end{aligned}

(b)

Write, in Cartesian components of unit vectors \hat{\mathbf{i}} and \hat{\mathbf{j}}, the following:

\begin{aligned} \mathbf{AB} & = |\mathbf{AB}|\cos 30^\circ\,\hat{\mathbf{i}} + |\mathbf{AB}|\sin 30^\circ\,\hat{\mathbf{j}} \\ & = 2\cos 30^\circ\,\hat{\mathbf{i}} + 2\sin 30^\circ\,\hat{\mathbf{j}} \\ & = \sqrt{3}\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \end{aligned}

and similarly,

\begin{aligned} \mathbf{u} & = |\mathbf{u}|\,\hat{\mathbf{i}} \\ \mathbf{v} & = |\mathbf{v}|\cos 60^\circ\,\hat{\mathbf{i}} + |\mathbf{v}|\sin 60^\circ\,\hat{\mathbf{j}}\\ & \\ \because\enspace & \mathbf{u}, \mathbf{v}\textrm{ are unit vectors} \\ \therefore\enspace & |\mathbf{u}|=|\mathbf{v}|=1 \\ & \\ \mathbf{u} & = \hat{\mathbf{i}} \\ \mathbf{v} & = \frac{1}{2}\,\hat{\mathbf{i}} + \frac{\sqrt{3}}{2} \hat{\mathbf{j}} \\ \end{aligned}

\begin{aligned} \begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}^{-1}\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} \\ \end{aligned}

Lemma. (Inversion of 2-by-2 matrices)

For 2\times 2 matrices, inversion can be done as follows:

\begin{aligned} \mathbf{A}^{-1} & = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} \\ & = \frac{1}{\mathrm{det}\mathbf{A}}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ & = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ \end{aligned}

Wikipedia on Invertible matrix

Then,

\begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}

so that

\begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix}

Thus

\begin{aligned} \mathbf{AB} & =\begin{pmatrix} \sqrt{3} & 1\end{pmatrix}\begin{bmatrix}\hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ & = \begin{pmatrix} \sqrt{3} & 1\end{pmatrix} \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix} \begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ & = \begin{bmatrix} \frac{2}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ &=\frac{2}{\sqrt{3}}\,\mathbf{u}+\frac{2}{\sqrt{3}}\,\mathbf{v} \\ \end{aligned}

There might be some mistake if conceptually.

(c)

I don’t know how. Skip it.