Vector Algebra
Parallelogram Law / Diagonals
Grade 12
Question:
<p>In a parallelogram \(ABCD\), let \(\overrightarrow{AB} = \vec{a}\) and \(\overrightarrow{AD} = \vec{b}\) (as shown in figure). Given \(|\vec{a}| = 2\), \(|\vec{b}| = 2\), and \(\vec{a} \cdot \vec{b} = \frac{1}{2}\). The diagonal \(\overrightarrow{AC} = \vec{a} + \vec{b}\). If \(\vec{a} = 3\vec{\alpha} - \vec{\beta}\) and \(\vec{b} = \vec{\alpha} + 3\vec{\beta}\), find \(|\overrightarrow{AC}|\) and \(|\overrightarrow{BD}|\).</p>
<p>\(|\overrightarrow{AC}| = 4\sqrt{7}\)</p>
<p>\(|\overrightarrow{BD}| = \sqrt{10}\)</p>
<p>\(|\overrightarrow{AC}| = 4\sqrt{5}\)</p>
<p>\(|\overrightarrow{BD}| = 4\sqrt{7}\)</p>
Step-by-Step Solution
Key Concept: In a parallelogram, the diagonals are AC = a + b and BD = b - a. Use the dot product formula |v|² = v·v and the given constraint a·b = 1/2 to find magnitudes without explicitly solving for α and β.
Step 1: Find |AC| The diagonal AC = a + b. Compute |AC|^2: |AC|^2 = (a + b)·(a + b) = |a|^2 + 2(a·b) + |b|^2 = 2^2 + 2(1/2) + 2^2 = 4 + 1 + 4 = 9 ∴ |AC| = 3 Step 2: Find |BD| The diagonal BD = AD - AB = b - a. Compute |BD|^2: |BD|^2 = (b - a)·(b - a) = |b|^2 - 2(a·b) + |a|^2 = 2^2 - 2(1/2) + 2^2 = 4 - 1 + 4 = 7 ∴ |BD| = √7 Note: The expressions a = 3α - β and b = α + 3β are provided as context but are unnecessary for solving; they may be used to verify consistency of the given conditions. ∴ Answer: A (|AC| = 3, |BD| = √7)
Correct Answer: A