TS EAMCET · Maths · Vector Algebra
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 :
- 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 \(\frac{1}{2}\{(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{a}}+(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{b}}+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{c}}\}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}\}\)
Step-by-step Solution
Detailed explanation
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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