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2014 WAEC Further Mathematics Theory The position vectors of P, Q and R are \(11i + j, 5i + \frac{13}{3}j\)...

Further Mathematics
WAEC 2014

The position vectors of P, Q and R are \(11i + j, 5i + \frac{13}{3}j\) and \(2i + 6j\) respectively. 

(a) Show that P, Q and R lie on a straight line.

(b) Find the ratio of \(|\overrightarrow{PQ}| : |\overrightarrow{QR}|\)

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Explanation

(a) For P, Q and R to lie on a straight line,the gradient of any two of PQ, PR and QR must be equal.

Gradient of PR : \(\frac{6 - 1}{2 - 11} = -\frac{5}{9}\)

Gradient of QR : \(\frac{6 - \frac{13}{3}}{2 - 5} = \frac{\frac{5}{3}}{-3}\)

= \(- \frac{5}{9}\)

The gradients are the same therefore P, Q and R are on the same line.

(b) \(\overrightarrow{PQ} = (5i + \frac{13}{3}j) - (11i + j)\)

= \(-6i + \frac{10}{3}j\)

\(|\overrightarrow{PQ}| = \sqrt{(-6)^{2} + (\frac{10}{3})^{2}}\)

= \(\sqrt{36 + \frac{100}{9}}\)

= \(\sqrt{\frac{424}{9}}\)

\(\overrightarrow{QR} = (2i + 6j) - (5i + \frac{13}{3}j)\)

= \(-3i + \frac{5}{3}j\)

\(|\overrightarrow{QR}| = \sqrt{(-3)^{2} + (\frac{5}{3})^{2}}\)

= \(\sqrt{9 + \frac{25}{9}}\)

= \(\sqrt{\frac{106}{9}}\)

\(\frac{|\overrightarrow{PQ}|}{|\overrightarrow{QR}|} = \frac{\sqrt{\frac{424}{9}}}{\sqrt{\frac{106}{9}}}\)

= \(\frac{4}{1}\)

\(\therefore |\overrightarrow{PQ}| : |\overrightarrow{QR}| = 4 : 1\) 


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