Let overrightarrow OA =overrightarrow a , overrightarrow OB =12 overrightarrow a +4 overrightarrow b and overrightarrow OC =overrightarrow b , where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then frac area\; of\; the\; quadrilateral\; OABC area\; of \; S is equal to

Mathematics · Vector Algebra

JEE Main 2024 — 29 January, Shift 2 — Question 8

Let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{~b}}$ and $\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{b}}$ , where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $\frac{\text { area\; of\; the\; quadrilateral\; } \mathrm{OABC}}{\text { area\; of \;} \mathrm{S}}$ is equal to $\qquad$

Option A:
6
Option B:
10
Option C:
7
Option D:
8
Correct

Answer: D

Step-by-step solution

Area of parallelogram, $S=|\vec{a} \times \vec{b}|$

Area of quadrilateral $=\operatorname{Area}(\Delta \mathrm{OAB})+\operatorname{Area}(\Delta \mathrm{OBC})$ $=\frac{1}{2}\{|\overrightarrow{\mathrm{a}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{~b}})|+|\overrightarrow{\mathrm{b}} \times(12 \overrightarrow{\mathrm{a}}+4 \overrightarrow{\mathrm{~b}})|\}$ $=8|(\vec{a} \times \vec{b})|$

Ratio $=\frac{8|(\vec{a} \times \vec{b})|}{|(\vec{a} \times \vec{b})|}=8$

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Mathematics
Chapter: Vector Algebra
Topic: Vector or Cross Product of Two Vectors

Answer: D

Step-by-step solution

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