JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\overrightarrow{\mathrm{OA}}=2 \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{OC}}=3 \overrightarrow{\mathrm{b}}\), where \(O\) is the origin. If the area of the parallelogram with adjacent sides \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OC}}\) is \(15\) sq. units, then the area (in sq. units) of the quadrilateral \(\mathrm{OABC}\) is equal to :
- A \(38\)
- B \(40\)
- C \(32\)
- D \(35\)
Answer & Solution
Correct Answer
(D) \(35\)
Step-by-step Solution
Detailed explanation
Area of parallelogram having sides \( \overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OC}}=|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OC}}|=|2 \overrightarrow{\mathrm{a}} \times 3 \overrightarrow{\mathrm{b}}|=15 \)…
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