Exam-Style Problem

0606 P21 - Jun 2023 - Q6 - 10 marks

7688

(a) The position vectors of the points \(P\), \(Q\) and \(R\), relative to an origin \(O\), are

\(\begin{pmatrix}4\\7\end{pmatrix},\quad \begin{pmatrix}8\\5\end{pmatrix},\quad \begin{pmatrix}x\\y\end{pmatrix}\)

respectively. The point \(R\) lies on \(PQ\) extended such that \(3\overrightarrow{QR}=2\overrightarrow{PR}\). Use a vector method to find the values of \(x\) and \(y\).

(b) You are given that \(\mathbf i\) is a unit vector due east and \(\mathbf j\) is a unit vector due north.

Three vectors, \(\mathbf a\), \(\mathbf b\) and \(\mathbf c\), are in the same horizontal plane as \(\mathbf i\) and \(\mathbf j\), and are such that \(\mathbf a+\mathbf b=\mathbf c\). The magnitude and bearing of \(\mathbf a\) are \(5\) and \(210^\circ\). The magnitude and bearing of \(\mathbf c\) are \(10\) and \(330^\circ\).

(i) Find \(\mathbf a\) and \(\mathbf c\) in terms of \(\mathbf i\) and \(\mathbf j\).

(ii) Find the magnitude and bearing of \(\mathbf b\).

Solution

Answer: (a) \(x=16\), \(y=1\); (b)(i) \(\mathbf a=-\frac52\mathbf i-\frac{5\sqrt3}{2}\mathbf j\), \(\mathbf c=-5\mathbf i+5\sqrt3\mathbf j\); (b)(ii) \(|\mathbf b|=5\sqrt7\approx13.2\), bearing \(349^\circ\).

Write each position or direction vector in terms of the given basis vectors, then compare coefficients to find the required constants or ratio.

(a) We have

\(\overrightarrow{QR}= \begin{pmatrix}x-8\\y-5\end{pmatrix}, \qquad \overrightarrow{PR}= \begin{pmatrix}x-4\\y-7\end{pmatrix}.\)

Since \(3\overrightarrow{QR}=2\overrightarrow{PR}\),

\(3\begin{pmatrix}x-8\\y-5\end{pmatrix} = 2\begin{pmatrix}x-4\\y-7\end{pmatrix}.\)

Equating components gives

\(3x-24=2x-8\)

and

\(3y-15=2y-14.\)

Therefore

\(x=16,\qquad y=1.\)

(b)(i) A bearing is measured clockwise from north. Since \(\mathbf i\) is east and \(\mathbf j\) is north, the east component is \(R\sin\theta\) and the north component is \(R\cos\theta\).

For \(\mathbf a\), \(R=5\) and \(\theta=210^\circ\), so

\(\mathbf a=5\sin210^\circ\,\mathbf i+5\cos210^\circ\,\mathbf j.\)

Hence

\(\mathbf a=-\frac52\mathbf i-\frac{5\sqrt3}{2}\mathbf j.\)

For \(\mathbf c\), \(R=10\) and \(\theta=330^\circ\), so

\(\mathbf c=10\sin330^\circ\,\mathbf i+10\cos330^\circ\,\mathbf j.\)

Hence

\(\mathbf c=-5\mathbf i+5\sqrt3\mathbf j.\)

(b)(ii) Since \(\mathbf a+\mathbf b=\mathbf c\),

\(\mathbf b=\mathbf c-\mathbf a.\)

Therefore

\(\mathbf b= \left(-5+\frac52\right)\mathbf i +\left(5\sqrt3+\frac{5\sqrt3}{2}\right)\mathbf j.\)

\(\mathbf b=-\frac52\mathbf i+\frac{15\sqrt3}{2}\mathbf j.\)

The magnitude is

\(|\mathbf b| =\sqrt{\left(-\frac52\right)^2+\left(\frac{15\sqrt3}{2}\right)^2} =\sqrt{\frac{25}{4}+\frac{675}{4}} =\sqrt{175} =5\sqrt7.\)

Thus

\(|\mathbf b|\approx13.2.\)

The vector is in the north-west direction. The angle west of north is

\(\tan^{-1}\left(\frac{2.5}{15\sqrt3/2}\right)\approx10.9^\circ.\)

Therefore the bearing is

\(360^\circ-10.9^\circ\approx349^\circ.\)