#### Need solution for RD sharma maths class 12 chapter 24 Vector or cross product exercise Fill in the blanks question 8

Answer: Area of triangle ABC = 10 sq. units

Hint: For any triangle, Area=
$\frac{1}{2}=\left | \vec{AB} \times \vec{AC} \right |$

Given:

$\left | \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}\right |=20$

As shown in figure let, O be the origin &triangle forming between A, B & C.

Let,

$OA=\vec{a}\\ OB=\vec{b}\\ OC=\vec{c}$

Now,

$\vec{AB}$= Position of B - Position of A
$=OB-OA\\ =\vec{b}-\vec{a}$

Similarly
$\vec{AC}$= Position of C - Position of A

$=OC-OA\\ =\vec{c}-\vec{a}$

Now, Area of

$\bigtriangleup ABC=\frac{1}{2}\left | \vec{AB} \times \vec{AC}\right |\\ \vec{AB} \times \vec{AC}=(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})$

$=(\vec{b} \times \vec{c}) -(\vec{b} \times \vec{a})-(\vec{a} \times \vec{c})+(\vec{a} \times \vec{a})\\ =(\vec{b} \times \vec{c}) -(\vec{b} \times \vec{a})-(\vec{a} \times \vec{c})+(0)\\$

$\left (\because (\vec{a} \times \vec{b})= -(\vec{b} \times \vec{a}) \right )\\ \left (\because (\vec{a} \times \vec{a})= 0 \right )\\$

$=\left [(\vec{b} \times \vec{c}) +(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a}) \right ]$............(1)

Using (1)

Area of

$\bigtriangleup ABC =\frac{1}{2} \times \left [(\vec{b} \times \vec{c}) +(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a}) \right ]$

$=\frac{1}{2} \times 20$

=10 sq. units