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 angle between $\vec a$ and $\vec b$ is $\frac{\pi}{4}$, is:
Answer & explanation
Correct answer: option 3
If angle between $\vec a$ and $\vec b$ is acute, then longer diagonal is at the common vertex.
However, if the angle between them is obtuse, then the longer diagonal is $\vec a -\vec b$ (another diagonal)
As two side are: $5\vec a+2\vec b,\,\vec a+3\vec b$
We have, $(5\vec a+2\vec b).(\vec a-3\vec b)=5×8-6×9-13\vec a.\vec b=40-54-12×\sqrt{2}×3×\frac{1}{\sqrt{2}}<0$
⇒ The angle is obtuse.
Now, longer diagonal is given as: $(5\vec a+2\vec b)-(\vec a-3\vec b)=4\vec a+5\vec b$
The length of longer diagonal is $|4\vec a+5\vec b|$
$=\sqrt{16×9+25×9+40\vec a.\vec b}=\sqrt{128+225+40×2\sqrt{2}×3×\frac{1}{\sqrt{2}}}=\sqrt{593}units$