Four vectors \(r = \alpha i + \beta j\), where... | WAEC SSCE - Further Mathematics - 2016

Question 1 Report

Four vectors \(r = \alpha i + \beta j\), where \(\alpha \text{ and } \beta\) are constants, \(s = 2i -j, m = 3i + 2j\) and \(n = i + j\) are such that the magnitude of r is three times as s and is parallel to the vactor (m - n).

(a) Find the values of \(\alpha\) and \(\beta\).

(b) Calculate the magnitude and direction of (r - s).

Answer

(a) Since \(r\) is parallel to \(m - n\):

\[m - n = (3i + 2j) - (i + j) = 2i + j\]

So \(r = \lambda(2i + j)\), giving \(\alpha = 2\lambda,\ \beta = \lambda\).

The magnitude of \(r\) is three times that of \(s = 2i - j\), where \(|s| = \sqrt{4 + 1} = \sqrt{5}\):

\[|r| = \sqrt{(2\lambda)^2 + \lambda^2} = |\lambda|\sqrt{5} = 3\sqrt{5} \ \Rightarrow\ |\lambda| = 3\]

Taking \(\lambda = 3\): \(\alpha = 6,\ \beta = 3\) (the other solution \(\lambda = -3\) gives \(\alpha = -6,\ \beta = -3\)).

(b) Using \(r = 6i + 3j\):

\[r - s = (6 - 2)i + (3 - (-1))j = 4i + 4j\] \[|r - s| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \approx 5.7\]

Direction: \(\tan\theta = \dfrac{4}{4} = 1\), so \(\theta = 45^\circ\) above the positive \(x\)-axis (bearing \(045^\circ\)).

Read lesson note on Vectors (WAEC)

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