JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}\), and \(\overrightarrow{O C}=\vec{b}\), where \(O\) is the origin. If \(S\) is the parallelogram with adjacent sides \(O A\) and \(O C\), then find the value of \(\frac{\text { area of quadrilateral } O A B C}{\text { area of } S} .\)
- A \(6\)
- B \(10\)
- C \(7\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\( \text { Area of parallelogram, } S=|\vec{a} \times \vec{b}| \) \( \text { Area of quadrilateral }=\text { Area }(\triangle \mathrm{OAB})+\text { Area }(\triangle \mathrm{OBC}) \)…
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