Correct Answer - Option 2 :
\(\rm \dfrac{1}{2} |\vec{a}\times \vec{b} + \vec{b} \times \vec{c}+\vec{c}\times \vec{a}|\)
Concept:
If the position vectors of points A and B are \(\rm \vec {a}\) and \(\rm \vec {b}\) respectively, then \(\rm \vec {AB}=\vec b - \vec a\).
Area of a Triangle with vectors \(\rm \vec {AB}\) and \(\rm \vec {AC}\) as its sides is given by: \(\rm Area=\dfrac12|\vec{A}\times\vec{B}|\).
Cross Product: For two vectors \(\rm \vec {A}=a_1\hat i+a_2\hat j+a_3\hat k\) and \(\rm \vec {B}=b_1\hat i+b_2\hat j+b_3\hat k\), their cross product is given by:
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\(\rm \vec A \times \vec B=\begin{vmatrix} \rm \hat i & \rm \hat j & \rm \hat k\\ \rm a_1& \rm a_2 & \rm a_3\\ \rm b_1 & \rm b_2 & \rm b_3\end{vmatrix}=(a_2b_3-a_3b_2)\hat i+(a_3b_1-a_1b_3)\hat j+(a_1b_2-a_2b_1)\hat k\).
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\(\rm \vec A\times\vec B=-\left(\vec B \times \vec A\right)\).
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\(\rm \vec A\times\vec A=0\).
The magnitude \(\rm |\vec A|\) of a vector \(\rm \vec {A}=a_1\hat i+a_2\hat j+a_3\hat k\) is given by: \(\rm |\vec A|=\sqrt{a_1^2+a_2^2+a_3^2}\).
Calculation:
We have \(\rm \vec {AB}=\vec b - \vec a\) and \(\rm \vec {AC}=\vec a - \vec c\).
Using the formula for the area of a triangle:
\(\rm Area=\dfrac{1}{2}\left|\vec {AB}\times\vec {AC}\right|\)
= \(\rm \dfrac{1}{2}\left|\left(\vec b-\vec a\right)\times\left(\vec c-\vec a\right)\right|\)
= \(\rm \dfrac{1}{2}\left|\vec b\times\vec c-\vec b\times\vec a -\vec a\times\vec c+\vec a\times\vec a\right|\)
= \(\rm \dfrac{1}{2} |\vec{a}\times \vec{b} + \vec{b} \times \vec{c}+\vec{c}\times \vec{a}|\).
Area of a parallelogram with vectors \(\rm \vec {a}\) and \(\rm \vec {b}\) as its sides is given by: \(\rm Area=\left|\vec{a}\times\vec{b}\right|\).
For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:
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Dot Product is defined as: \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
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Cross Product is defined as: \(\rm \vec A\times \vec B=\hat n|\vec A||\vec B|\sin \theta\), where \(\rm \hat n\) is the unit vector perpendicular to the plane containing \(\rm \vec A\) and \(\rm \vec B\).
