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

Answers (1)

Answer: Area of triangle ABC = 10 sq. units

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


\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.


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


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

\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

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