Let OA = a, OB = 12a + 4b, and OC = b, where O is the origin. If… — Solution | JEE Mathematics

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

A
$6$
B
$10$
C
$7$
D
$8$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $\overrightarrow{OA}=\vec{a}$ , $\overrightarrow{OB}=12\vec{a}+4\vec{b}$ , and $\overrightarrow{OC}=\vec{b}$ .

Find: The ratio of the area of quadrilateral $OABC$ to the area of parallelogram $S$ .

The area of parallelogram $S$ with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is

\text{Area}(S)=|\vec{a}\times \vec{b}|

Split quadrilateral $OABC$ into triangles $\triangle OAB$ and $\triangle OBC$ .

For $\triangle OAB$ ,

\text{Area}(\triangle OAB)=\frac{1}{2}|\overrightarrow{OA}\times \overrightarrow{OB}|

Now,

\overrightarrow{OA}\times \overrightarrow{OB}=\vec{a}\times (12\vec{a}+4\vec{b})=4(\vec{a}\times \vec{b})

So,

\text{Area}(\triangle OAB)=\frac{1}{2}|4\vec{a}\times \vec{b}|=2|\vec{a}\times \vec{b}|

For $\triangle OBC$ ,

\text{Area}(\triangle OBC)=\frac{1}{2}|\overrightarrow{OB}\times \overrightarrow{OC}|

Now,

\overrightarrow{OB}\times \overrightarrow{OC}=(12\vec{a}+4\vec{b})\times \vec{b}=12(\vec{a}\times \vec{b})

because $\vec{b}\times \vec{b}=0$ . Hence,

\text{Area}(\triangle OBC)=\frac{1}{2}|12\vec{a}\times \vec{b}|=6|\vec{a}\times \vec{b}|

Therefore, the area of quadrilateral $OABC$ is

\text{Area}(OABC)=2|\vec{a}\times \vec{b}|+6|\vec{a}\times \vec{b}|=8|\vec{a}\times \vec{b}|

Hence, the required ratio is

\frac{\text{Area}(OABC)}{\text{Area}(S)}=\frac{8|\vec{a}\times \vec{b}|}{|\vec{a}\times \vec{b}|}=8

Therefore, the correct option is D.

Direct Area Comparison

Given: $\overrightarrow{OA}=\vec{a}$ , $\overrightarrow{OB}=12\vec{a}+4\vec{b}$ , and $\overrightarrow{OC}=\vec{b}$ .

Find: $\dfrac{\text{Area}(OABC)}{\text{Area}(S)}$ .

Use the fact that every required area is a multiple of $|\vec{a}\times \vec{b}|$ .

Since

\text{Area}(S)=|\vec{a}\times \vec{b}|

and

\text{Area}(OABC)=\frac{1}{2}|\vec{a}\times (12\vec{a}+4\vec{b})|+\frac{1}{2}|(12\vec{a}+4\vec{b})\times \vec{b}|

we get

\text{Area}(OABC)=\frac{1}{2}|4\vec{a}\times \vec{b}|+\frac{1}{2}|12\vec{a}\times \vec{b}|=2|\vec{a}\times \vec{b}|+6|\vec{a}\times \vec{b}|=8|\vec{a}\times \vec{b}|

Thus,

\frac{\text{Area}(OABC)}{\text{Area}(S)}=\frac{8|\vec{a}\times \vec{b}|}{|\vec{a}\times \vec{b}|}=8

Therefore, the correct option is D.

Common mistakes

Students often use dot product instead of cross product for area. This is wrong because area of a parallelogram formed by two vectors is given by the magnitude of their cross product. Use $|\vec{a}\times \vec{b}|$ , not $\vec{a}\cdot\vec{b}$ .
A common mistake is to treat quadrilateral $OABC$ as a parallelogram directly. This is incorrect because the given vertices do not imply opposite sides are parallel. Split it into $\triangle OAB$ and $\triangle OBC$ and add the two areas.
Many students forget that $\vec{a}\times \vec{a}=0$ and $\vec{b}\times \vec{b}=0$ . Because of this, they incorrectly keep extra terms while expanding $\vec{a}\times (12\vec{a}+4\vec{b})$ or $(12\vec{a}+4\vec{b})\times \vec{b}$ . Remove the self-cross-product terms.

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Answer

Step-by-step solution

Common mistakes

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