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Question: Answered & Verified by Expert
Let $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle A B C$. The vector area of $\triangle A B C$ is :
MathematicsVector AlgebraTS EAMCETTS EAMCET 2003
Options:
  • A $\frac{1}{2}\{\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})+\overrightarrow{\mathbf{b}} \times(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})+\overrightarrow{\mathbf{c}} \times(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})\}$
  • B $\frac{1}{2}\{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}\}$
  • C $\frac{1}{2}\{\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\}$
  • D $\frac{1}{2}\{(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{a}}+(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{b}}+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{c}}\}$
Solution:
1620 Upvotes Verified Answer
The correct answer is: $\frac{1}{2}\{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}\}$
We have,
Area of the triangle
$=\frac{1}{2}\{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}\}$

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