Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be vectors, satisfying \[\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|=5\] and $\left( \overrightarrow{a},\overrightarrow{b} \right)={{45}^{\circ }}$. Find the area of the triangle having $\overrightarrow{a}-2\overrightarrow{b}$and $3\overrightarrow{a}+2\overrightarrow{b}$as two of its sides.

Hint: We can define the area of a triangle as the region that is enclosed by the three sides. We can calculate the area of the triangle by knowing the base length and height. Or in other words if we know the two sides of a triangle the half the product of the two sides gives the area which is a vector quantity.

Complete step-by-step solution:
Area of triangle $=\dfrac{1}{2}\left[ \left( \overrightarrow{a}-2\overrightarrow{b} \right)\times \left( 3\overrightarrow{a}+2\overrightarrow{b} \right) \right]$
                 $=\dfrac{1}{2}\left[ \left( \overrightarrow{a}\times 2\overrightarrow{b} \right)-\left( 2\overrightarrow{b}\times 3\overrightarrow{a} \right) \right]$
                 $=\dfrac{1}{2}\left[ 2ab\sin \theta -6ba\sin \theta \right]$
Given that the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is 45 degrees.
Substitute this in the above equation we get,
$A=\dfrac{1}{2}\left[ 2ab\sin {{45}^{\circ }}-6ba\sin {{45}^{\circ }} \right]$
$A=\dfrac{1}{2}\left[ 2a{{b}^{{}}}\times \dfrac{1}{\sqrt{2}}-6ba\times \dfrac{1}{\sqrt{2}} \right]$
$A=\dfrac{1}{\sqrt{2}}\left[ ab-3ba \right]$
$A=\dfrac{1}{\sqrt{2}}\left[ -2ab \right]$
Since the area is a positive quantity, here we consider only the magnitude.
Thus the area of the triangle is,
$A=\sqrt{2}\left| ab \right|$
$A=\sqrt{2}\times 5\times 5$
$\therefore A=25\sqrt{2}$

Note: In vector algebra we can see various types of vectors. We can also perform various operations like addition, subtraction, product etc. using vectors. There are some rules assigned with. We can thus define the cross product of a vector as a binary operation. Usually we consider a three dimensional space. The cross product of two vectors gives another vector and its direction will be perpendicular to both the vectors. The vector product is otherwise known as the cross products. For calculating the length of a vector or to find the angle between them we use dot product. And also the cross product is not commutative. The right hand thumb rule is used to find the direction of a unit vector.