Given |vec a|=5 , |vec b|=5 , angle(vec a,vec b)=dfracπ4…

Given $|\vec a|=5$, $|\vec b|=5$, $\angle(\vec a,\vec b)=\dfrac\pi4$, find the area of the parallelogram constructed on $2\vec b-\vec a$ and $3\vec a+2\vec b$.

Explanation of Method

The area of the parallelogram on 2b−a and 3a+2b.

Step 1: Expansion

\[ (2\vec b-\vec a)\times(3\vec a+2\vec b)=6\,\vec b\times\vec a+4\,\vec b\times\vec b-3\,\vec a\times\vec a-2\,\vec a\times\vec b=-8\,\vec a\times\vec b. \]

(With the rule from Problem 94: $2\vec b-\vec a=-\vec a+2\vec b$ and $3\vec a+2\vec b$ give $(-1)\cdot2-2\cdot3=-8$.)

Step 2: Area

\[ |\vec a\times\vec b|=5\cdot5\cdot\frac{\sqrt2}{2}=\frac{25\sqrt2}{2},\qquad S=8\cdot\frac{25\sqrt2}{2}. \]

\[S=100\sqrt2\approx141.42\]

$O$ $\vec a$ $\vec b$ $2\vec b-\vec a$ $3\vec a+2\vec b$

Drawn to scale: the orange parallelogram has $8$ times the area of the blue one ($\tfrac{25\sqrt2}{2}$), i.e. $100\sqrt2$.

Result

\[\boxed{\,S=100\sqrt2\approx141.42\,}\]