Question

Let \( \vec{a}, \vec{b} \) be non-collinear vectors, with \( |\vec{a}| = 2\sqrt{2} \), \( |\vec{b}| = 3 \), and the angle between them is \( 45^\circ \). Then the lengths of the diagonals of the parallelogram whose adjacent sides are \( \vec{5a} + \vec{2b} \) and \( \vec{a} - \vec{3b} \) are:

• A. \( 15, 593 \)
• B. \( 15, \sqrt{593} \)
• C. \( 225, \sqrt{593} \)
• D. \( 225, 593 \)

Hint:

Diagonals of a Parallelogram}
Use \( \vec{u} + \vec{v} \), \( \vec{u} - \vec{v} \) to find diagonals
Use dot product \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \)
Calculate square of magnitude and simplify

Answer 1

The Correct Option is B

Let \( \vec{u} = \vec{5a} + \vec{2b} \), \( \vec{v} = \vec{a} - \vec{3b} \) The diagonals of the parallelogram are \( \vec{u} + \vec{v} \) and \( \vec{u} - \vec{v} \) \[ \vec{u} + \vec{v} = 6\vec{a} - \vec{b}, \quad \vec{u} - \vec{v} = 4\vec{a} + 5\vec{b} \] Now calculate magnitudes using vector identities: \[ |\vec{a}| = 2\sqrt{2}, \quad |\vec{b}| = 3, \quad \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(45^\circ) = 2\sqrt{2} \cdot 3 \cdot \frac{1}{\sqrt{2}} = 6 \] Calculate \( |\vec{u} + \vec{v}| = |6\vec{a} - \vec{b}| \): \[ |6\vec{a} - \vec{b}|^2 = 36|\vec{a}|^2 + |\vec{b}|^2 - 2 \cdot 6 \cdot \vec{a} \cdot \vec{b} = 36 \cdot 8 + 9 - 72 = 225 \Rightarrow \text{Length} = 15 \] \[ |4\vec{a} + 5\vec{b}|^2 = 16|\vec{a}|^2 + 25|\vec{b}|^2 + 40(\vec{a} \cdot \vec{b}) = 128 + 225 + 240 = 593 \Rightarrow \text{Length} = \sqrt{593} \]