Vectors 3 Question 21
21. Let $\overrightarrow{\mathbf{O A}}=\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{O B}}=10 \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{O C}}=\overrightarrow{\mathbf{b}}$, where $O, A$ and $C$ are non-collinear points. Let $p$ denotes the area of the quadrilateral $O A B C$ and let $q$ denotes, the area of the parallelogram with $O A$ and $O C$ as adjacent sides. If $p=k q$, then $k=$
(1997, 2M)
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Answer:
Correct Answer: 21. (6)
Solution:
- Since, $q=$ area of parallelogram with $\overrightarrow{O A}$ and $\overrightarrow{O C}$ as adjacent sides $=|\overrightarrow{\mathbf{O A}} \times \overrightarrow{\mathbf{O C}}|=|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
and $p=$ area of quadrilateral $O A B C$
$$ \begin{aligned} & =\frac{1}{2}|\overrightarrow{\mathbf{O A}} \times \overrightarrow{\mathbf{O B}}|+\frac{1}{2}|\overrightarrow{\mathbf{O B}} \times \overrightarrow{\mathbf{O C}}| \\ & =\frac{1}{2}\left|\overrightarrow{\mathbf{a}} \times(10 \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}})+\frac{1}{2}\right|(10 \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{b}} \mid \\ & =|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|+5|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=6|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}| \\ \therefore \quad p & =6 q \Rightarrow \quad k=6 \end{aligned} $$
