Given that \(r = 2i - j\), \(s = 3i + 5j\) and \(t = 6i - 2j\), find the magnitude of \(2r + s - t\).

Assessment: WAEC SSCE - Further Mathematics - 2015 (Objective) Subject: Further Mathematics

Question 1 Report

Given that \(r = 2i - j\), \(s = 3i + 5j\) and \(t = 6i - 2j\), find the magnitude of \(2r + s - t\).

Answer Details
We are given that: \begin{align*} r &= 2i - j\\ s &= 3i + 5j\\ t &= 6i - 2j \end{align*} We can now substitute these expressions into \(2r+s-t\) and simplify: \begin{align*} 2r + s - t &= 2(2i-j) + (3i+5j) - (6i - 2j)\\ &= 4i - 2j + 3i + 5j - 6i + 2j\\ &= i + 5j \end{align*} The magnitude of a vector with components \(a\) and \(b\) is given by \(\sqrt{a^2 + b^2}\). So, the magnitude of \(i+5j\) is: \begin{align*} \sqrt{i^2 + 5j^2} &= \sqrt{1^2 + 5^2} \\ &= \sqrt{26} \end{align*} Therefore, the answer is \(\sqrt{26}\).
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