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Let $\vec{O} A = \vec{a}, \vec{O} B = 10 \vec{a} + 2 \vec{b}$ and $\vec{O} C = \vec{b},$ where $O, A$ and $C$ are non-collinear points. Let p denote the area of quadrilateral $O A C B$ , and let q denote the area of parallelogram with $O A$ and $O C$ as adjacent sides. If p = kq , then find k.

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answer is 6.

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Detailed Solution

$Here \vec{O} A = \vec{a}, \vec{O} B = 10 \vec{a} + 2 \vec{b}, \vec{O} C = \vec{b}$

$∴ q = | \vec{a} \times \vec{b} | . . . . . . (i)$

$p =$ Area of quadrilateral $O A B C$

$= Area of △ O A B + Area of △ O B C$

$= \frac{1}{2} | \vec{a} \times (10 \vec{a} + 2 \vec{b}) | + \frac{1}{2} | (10 \vec{a} + 2 \vec{b}) \times \vec{b} | ∴ k = 6$

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