Question:
A parallelogram is constructed on the vectors $\mathbf{a} = 3\alpha - \beta$, $\mathbf{b} = \alpha + 3\beta$. If $|\alpha| = |\beta| = 2$ and the angle between $\alpha$ and $\beta$ is $\dfrac{\pi}{3}$, then the length of a diagonal of the parallelogram is
A parallelogram is constructed on the vectors $\mathbf{a} = 3\alpha - \beta$, $\mathbf{b} = \alpha + 3\beta$. If $|\alpha| = |\beta| = 2$ and the angle between $\alpha$ and $\beta$ is $\dfrac{\pi}{3}$, then the length of a diagonal of the parallelogram is
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The diagonals of a parallelogram are the sum and difference of its adjacent side vectors. Use the dot product formula $\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$ when the angle between vectors is given.
Updated On: May 2, 2026
- $4\sqrt{3}$
- $4\sqrt{5}$
- $4\sqrt{7}$
- None of these
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The diagonals of a parallelogram are given by the sum and difference of the adjacent side vectors.
Step 2: Detailed Explanation:
One diagonal $= \mathbf{a} + \mathbf{b} = (3\alpha - \beta) + (\alpha + 3\beta) = 4\alpha + 2\beta$.
\[ |\mathbf{a}+\mathbf{b}|^2 = 16|\alpha|^2 + 4|\beta|^2 + 2(4)(2)\,\alpha \cdot \beta = 16(4) + 4(4) + 16 \cdot |\alpha||\beta|\cos\tfrac{\pi}{3} = 64 + 16 + 16 \cdot 2 \cdot 2 \cdot \tfrac{1}{2} = 64 + 16 + 32 = 112 \] So $|\mathbf{a}+\mathbf{b}| = \sqrt{112} = 4\sqrt{7}$.
Step 3: Final Answer:
The length of the diagonal is $4\sqrt{7}$.
The diagonals of a parallelogram are given by the sum and difference of the adjacent side vectors.
Step 2: Detailed Explanation:
One diagonal $= \mathbf{a} + \mathbf{b} = (3\alpha - \beta) + (\alpha + 3\beta) = 4\alpha + 2\beta$.
\[ |\mathbf{a}+\mathbf{b}|^2 = 16|\alpha|^2 + 4|\beta|^2 + 2(4)(2)\,\alpha \cdot \beta = 16(4) + 4(4) + 16 \cdot |\alpha||\beta|\cos\tfrac{\pi}{3} = 64 + 16 + 16 \cdot 2 \cdot 2 \cdot \tfrac{1}{2} = 64 + 16 + 32 = 112 \] So $|\mathbf{a}+\mathbf{b}| = \sqrt{112} = 4\sqrt{7}$.
Step 3: Final Answer:
The length of the diagonal is $4\sqrt{7}$.
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