SATHEE: Vectors 3 Question 21

Vectors 3 Question 21

21. 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 $p$ 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 $\vec{O A}$ and $\vec{O C}$ as adjacent sides $= | \vec{O A} \times \vec{O C} | = | \vec{a} \times \vec{b} |$

and $p =$ area of quadrilateral $O A B C$

$\begin{aligned} = \frac{1}{2} | \vec{O A} \times \vec{O B} | + \frac{1}{2} | \vec{O B} \times \vec{O C} | \\ = \frac{1}{2} | \vec{a} \times (10 \vec{a} + 2 \vec{b}) + \frac{1}{2} | (10 \vec{a} + 2 \vec{b}) \times \vec{b} ∣ \\ = | \vec{a} \times \vec{b} | + 5 | \vec{a} \times \vec{b} | = 6 | \vec{a} \times \vec{b} | \\ ∴ p & = 6 q \Rightarrow k = 6 \end{aligned}$

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Vectors 3 Question 21

21. Let OA→=a→,OB→=10a→+2b→ and OC→=b→, where O,A and C are non-collinear points. Let p denotes the area of the quadrilateral OABC and let q denotes, the area of the parallelogram with OA and OC as adjacent sides. If p=kq, then k=

Answer:

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