Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $|\overrightarrow{\mathbf{a}}|=2,|\overrightarrow{\mathbf{b}}|=3$ and $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ are mutually
perpendicular, then the area of the triangle whose vertices are $\overrightarrow{\mathbf{0}}, \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}$ is
MathematicsVector AlgebraMHT CETMHT CET 2008
Options:
  • A 5
  • B 1
  • C 6
  • D 8
Solution:
2686 Upvotes Verified Answer
The correct answer is: 6
Let the position vectors of the points $A, B, C$ are
$$
\begin{array}{l}
\mathbf{0}, \mathbf{a}+\mathbf{b}, \mathbf{a}-\mathbf{b} \text { and } \theta=90^{\circ} . \\
\therefore \text { Area of triangle }=\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}| \\
\left.\quad=\frac{1}{2} \mid \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\right) \times(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \mid \\
=\frac{1}{2}|2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}| \\
=|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{a}}| \sin \theta \\
=3 \times 2 \sin 90^{\circ} \\
=6
\end{array}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.