(a) Given that \(x = 3i - j, y = 2i + kj\) and the cosine of the angle between x and y is \(\frac{\sqrt{5}}{5}\), find the values of the constant k. (b) In ...

(a) With \(x=3i-j\) and \(y=2i+kj\):

\[x\cdot y=(3)(2)+(-1)(k)=6-k,\quad |x|=\sqrt{3^2+(-1)^2}=\sqrt{10},\quad |y|=\sqrt{4+k^2}.\]

Given \(\cos\theta=\dfrac{\sqrt5}{5}=\dfrac{1}{\sqrt5}\):

\[\frac{6-k}{\sqrt{10}\,\sqrt{4+k^2}}=\frac{1}{\sqrt5}.\]

Cross-multiply and square:

\[\sqrt5\,(6-k)=\sqrt{10}\,\sqrt{4+k^2}\;\Rightarrow\;5(6-k)^2=10(4+k^2).\]

\[(6-k)^2=2(4+k^2)\;\Rightarrow\;36-12k+k^2=8+2k^2.\]

\[k^2+12k-28=0\;\Rightarrow\;(k-2)(k+14)=0.\]

So \(k=2\) or \(k=-14\). Both keep \(6-k>0\) (so \(\cos\theta>0\)), hence both are valid: \(\mathbf{k=2}\) or \(\mathbf{k=-14}\).

(b) Take \(A\) as the origin. Then, using the given vectors:

\[\overrightarrow{AB}=\begin{pmatrix}-5\\-1\end{pmatrix}\Rightarrow B=(-5,-1),\qquad \overrightarrow{AC}=\begin{pmatrix}-6\\-9\end{pmatrix}\Rightarrow C=(-6,-9).\]

\[D=B+\overrightarrow{BD}=\begin{pmatrix}-5\\-1\end{pmatrix}+\begin{pmatrix}4\\-7\end{pmatrix}=\begin{pmatrix}-1\\-8\end{pmatrix}\Rightarrow D=(-1,-8).\]

For ABCD (vertices in order) to be a parallelogram, opposite sides must be equal, i.e. \(\overrightarrow{AB}=\overrightarrow{DC}\). Now

\[\overrightarrow{DC}=C-D=\begin{pmatrix}-6-(-1)\\-9-(-8)\end{pmatrix}=\begin{pmatrix}-5\\-1\end{pmatrix}=\overrightarrow{AB}.\]

Since \(\overrightarrow{AB}=\overrightarrow{DC}\), the side AB is equal and parallel to DC. As a check, \(\overrightarrow{AD}=D-A=\begin{pmatrix}-1\\-8\end{pmatrix}\) and \(\overrightarrow{BC}=C-B=\begin{pmatrix}-1\\-8\end{pmatrix}\), so \(\overrightarrow{AD}=\overrightarrow{BC}\) as well. (Equivalently, the diagonals bisect each other: midpoint of \(AC=(-3,-4.5)=\) midpoint of \(BD\).)

Therefore ABCD is a parallelogram.