(a) Three vectors a, b and c are \(\begin{pmatrix} 8 \\ 3 \end{pmatrix}, \begin{pmatrix} 6 \\ -5 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) respectively. Find the vector d such that \(|d| = \sqrt{41}\) and d is in the direction of \(a + b – 2c\).
(b) The coordinates of A and B are (3, 4) and (3, n) respectively. If AOB = 30°, find, correct to 2 decimal places, the values of n.
Explanation
(a) \(a = \begin{pmatrix} 8 \\ 3 \end{pmatrix} ; b = \begin{pmatrix} 6 \\ -5 \end{pmatrix} ; c = \begin{pmatrix} 2 \\ -3 \end{pmatrix} ; |d| = \sqrt{41}\)
\(p = a + b - 2c\)
= \(\begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} + 2\begin{pmatrix} 2 \\ -3 \end{pmatrix}\)
= \(\begin{pmatrix} 8 + 6 + 4 \\ 3 - 5 - 6 \end{pmatrix}\)
= \(\begin{pmatrix} 10 \\ -8 \end{pmatrix}\)
\(|p| = \sqrt{10^{2} + (-8)^{2}} = \sqrt{164}\)
\(|p| = 2\sqrt{41}\)
\(|d| = \frac{1}{2} |p|\)
\(\therefore d = \frac{1}{2} p = \frac{1}{2} \begin{pmatrix} 10 \\ -8 \end{pmatrix}\)
= \(\begin{pmatrix} 5 \\ -4 \end{pmatrix}\)
(b) A(3, 4) ; B(3, n).
\(OA = 3i + 4j ; OB = 3i + nj\)
\(OA \cdot OB = |OA||OB| \cos \theta\)
\((3i + 4j) \cdot (3i + nj) = |3i + 4j||3i + nj| \cos 30\)
\(9 + 4n = 5 (\sqrt{9 + n^{2}}) (\sqrt{3}{2})\)
\(2(9 + 4n) = 5(\sqrt{3(9 + n^{2})})\)
\(18 + 8n = 5\sqrt{27 + 3n^{2}}\)
Squaring both sides,
\(324 + 288n + 64n^{2} = 25(27 + 3n^{2})\)
\(324 + 288n + 64n^{2} = 675 + 75n^{2}\)
\(75n^{2} - 64n^{2} - 288n + 675 - 324 = 0\)
\(11n^{2} - 288n + 351 = 0\)
\(n = \frac{-(-288) \pm \sqrt{(-288)^{2} - 4(11)(351)}}{2(11)}\)
\(n = \frac{288 \pm \sqrt{82944 - 15444}}{22}\)
\(n = \frac{288 \pm \sqrt{67500}}{22}\)
\(n = \frac{288 \pm 259.81}{22}\)
\(n = \frac{288 + 259.81}{22} ; n = \frac{288 - 259.81}{22}\)
\(n = \frac{547.81}{22} ; n = \frac{28.19}{22}\)
\(n = 24.9 ; n = 1.28\)