The length of the longer diagonal of the parallelogram constructed on $5\vec a+2\vec b$ and $\vec a-3\vec b$, if it is given that $|\vec a|=2\sqrt{2},|\vec b|=3$ and $\vec a.\vec b=\frac{π}{4}$, is
Answer & explanation
Correct answer: option 3
The diagonals of the parallelogram are
$\vec α=5\vec a+2\vec b+\vec a-3\vec b= 6\vec a-\vec b$ and $\vec β = ±(4\vec a+5\vec b)$
Now,
$|\vec α|=|6\vec a-\vec b|$
$⇒|\vec α|=\sqrt{36|\vec a|^2+|\vec b|^2-12 (\vec a.\vec b)}$
$⇒|\vec α|=\sqrt{36×8+9-12 × 2\sqrt{2} × 3×\frac{1}{\sqrt{2}}}=15$
and,
$|\vec β|=|4\vec a +5\vec b|$
$⇒|\vec β|=\sqrt{16|\vec a|^2+25|\vec b|^2+40(\vec a.\vec b)}$
$⇒|\vec β|=\sqrt{16×8+25×9+40×2\sqrt{2}×3×\frac{1}{\sqrt{2}}}=\sqrt{593}$
Clearly, $|\vec β|>|\vec α|$
Hence, the length of the longer diagonal is $\sqrt{593}$.