Vector Algebra
Magnitude of Vectors
Grade 12
Question:
<p><strong>Example 32.</strong> The length of longer diagonal of the parallelogram constructed on \(5\mathbf{a} + 2\mathbf{b}\) and \(\mathbf{a} - 3\mathbf{b}\), when it is given that \(|\mathbf{a}| = 2\sqrt{2}\), \(|\mathbf{b}| = 3\) and angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(\frac{\pi}{4}\), is</p>
<p>(a) 15</p>
<p>(b) 113</p>
<p>(c) \(\sqrt{593}\)</p>
<p>(d) \(\sqrt{369}\)</p>
Step-by-Step Solution
Key Concept: For a parallelogram with sides as vectors, the diagonals are the sum and difference of the side vectors. Use the magnitude formula with the given magnitudes and angle between vectors.
Step 1: The lengths of the two diagonals are: \(d_1 = |(5\mathbf{a} + 2\mathbf{b}) + (\mathbf{a} - 3\mathbf{b})| = |6\mathbf{a} - \mathbf{b}|\) \(d_2 = |(5\mathbf{a} + 2\mathbf{b}) - (\mathbf{a} - 3\mathbf{b})| = |4\mathbf{a} + 5\mathbf{b}|\) Step 2: Calculate \(d_2^2 = |4\mathbf{a} + 5\mathbf{b}|^2\) \(d_2^2 = 16|\mathbf{a}|^2 + 25|\mathbf{b}|^2 + 40|\mathbf{a}||\mathbf{b}|\cos\frac{\pi}{4}\) \(= 16 \times 8 + 25 \times 9 + 40 \times 2\sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}\) \(= 128 + 225 + 240 = 593\) Step 3: Therefore, length of the longer diagonal \(= \sqrt{593}\) ∴ Answer is C.
Correct Answer: C