Let OA = 2a, OB = 6a + 5b and OC = 3b, where O is the origin. If the area of the parallelogram with

Q: Wait, why do we split the quadrilateral into two triangles? Can't we just use a direct formula?

While there are coordinate-based formulas for the area of a quadrilateral, in vector geometry, decomposing complex shapes into simpler ones like triangles is the most robust and elegant method. It reduces the problem to the fundamental cross-product definition, minimizing the chance of algebraic errors.

Q: Why does the term $\vec{a} \times \vec{a}$ become zero?

Great question! The magnitude of the cross product is defined as $|\vec{u}| |\vec{v}| \sin(\theta)$, where $\theta$ is the angle between the vectors. If a vector is crossed with itself, the angle $\theta$ is $0^\circ$. Since $\sin(0^\circ) = 0$, the entire cross product vanishes. This is a powerful tool for simplifying vector expressions!

Q: Does the order of vectors in the cross product matter for the area?

The cross product is anti-commutative, meaning $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$. However, because we are taking the *magnitude* (the absolute value) of the cross product to find the area, the negative sign disappears. So, $|\vec{u} \times \vec{v}|$ is always equal to $|\vec{v} \times \vec{u}|$. You are safe either way!

Question

Let $\vec{OA} = 2\vec{a}, \vec{OB} = 6\vec{a} + 5\vec{b}$ and $\vec{OC} = 3\vec{b}$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\vec{OA}$ and $\vec{OC}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $OABC$ is equal to :

Options:

(A) 32

(B) 40

(C) 38

(D) 35

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into Area

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are painting a picture in vector space.

We have been given three vectors, $\vec{OA} = 2\vec{a}$, $\vec{OB} = 6\vec{a} + 5\vec{b}$, and $\vec{OC} = 3\vec{b}$, all originating from the origin $O$. Our mission is to find the area of the quadrilateral $OABC$.

This might seem daunting at first, but let us break it down into a beautiful, logical sequence.

### Phase 1: The Anchor of the Parallelogram

Every great journey begins with a solid foundation. We are told that the area of the parallelogram formed by $\vec{OA}$ and $\vec{OC}$ is $15$ square units.

In the language of vectors, the area of a parallelogram defined by two adjacent vectors $\vec{u}$ and $\vec{v}$ is simply the magnitude of their cross product, $|\vec{u} 	imes \vec{v}|$. So, we write:

$$|\vec{OA} 	imes \vec{OC}| = 15$$

Substituting our known values, we get $|2\vec{a} 	imes 3\vec{b}| = 15$. Using the scalar multiplication property of the cross product, we pull the constants $2$ and $3$ out:

$$6|\vec{a} 	imes \vec{b}| = 15$$

Dividing by $6$, we find our golden key: $|\vec{a} 	imes \vec{b}| = 2.5$. Keep this value close; it is the heartbeat of our entire calculation.

### Phase 2: The Art of Decomposition

Now, look at the quadrilateral $OABC$. It is not a standard shape, but geometry offers us a brilliant strategy: divide and conquer.

By drawing the diagonal $\vec{OB}$, we split the quadrilateral into two triangles: $	riangle OAB$ and $	riangle OBC$. The total area is simply the sum of the areas of these two triangles:

$$	ext{Area}(OABC) = 	ext{Area}(	riangle OAB) + 	ext{Area}(	riangle OBC)$$

### Phase 3: The Algebraic Dance

Let us calculate the area of $	riangle OAB$ first. The area of a triangle is half the area of the parallelogram formed by the same sides:

$$	ext{Area}(	riangle OAB) = \frac{1}{2} |\vec{OA} 	imes \vec{OB}|$$

Substituting $\vec{OA} = 2\vec{a}$ and $\vec{OB} = 6\vec{a} + 5\vec{b}$, we get:

$$\frac{1}{2} |2\vec{a} 	imes (6\vec{a} + 5\vec{b})| = \frac{1}{2} |12(\vec{a} 	imes \vec{a}) + 10(\vec{a} 	imes \vec{b})|$$

Here is the magic: $\vec{a} 	imes \vec{a} = 0$. The first term vanishes, leaving us with:

$$\frac{1}{2} |10(\vec{a} 	imes \vec{b})| = 5|\vec{a} 	imes \vec{b}|$$

Next, we tackle $	riangle OBC$. Its area is $\frac{1}{2} |\vec{OB} 	imes \vec{OC}|$. Substituting $\vec{OB} = 6\vec{a} + 5\vec{b}$ and $\vec{OC} = 3\vec{b}$:

$$\frac{1}{2} |(6\vec{a} + 5\vec{b}) 	imes 3\vec{b}| = \frac{1}{2} |18(\vec{a} 	imes \vec{b}) + 15(\vec{b} 	imes \vec{b})|$$

Since $\vec{b} 	imes \vec{b} = 0$, the second term vanishes, leaving us with:

$$\frac{1}{2} |18(\vec{a} 	imes \vec{b})| = 9|\vec{a} 	imes \vec{b}|$$

### Phase 4: The Final Synthesis

We are almost there! The total area is $5|\vec{a} 	imes \vec{b}| + 9|\vec{a} 	imes \vec{b}| = 14|\vec{a} 	imes \vec{b}|$.

Remember our golden key from Phase 1? We found that $|\vec{a} 	imes \vec{b}| = 2.5$. Substituting this in:

$$14 	imes 2.5 = 35$$

And there it is! The area of the quadrilateral $OABC$ is **$35$ square units**. You have successfully navigated the vector space, decomposed the geometry, and mastered the algebra. Well done!

Animated Solution for Mathematics - Vector Algebra: Let $O A = 2 a, O B = 6 a + 5 b$ and $O C = 3 b$ , where $O$ is the origin. If the area of the parallelogram with adjacent sides $O A$ and $O C$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $O A B C$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Base Vectors

Area of a Parallelogram

Substituting the Vectors

Extracting Constants

Finding $∣ a \times b ∣$

Introducing Vector $O B$

Splitting the Quadrilateral

Area of $△ O A B$ Setup

Calculating Area of $△ O A B$

Simplifying Area of $△ O A B$

Area of $△ O B C$ Setup

Calculating Area of $△ O B C$

Total Area Expression

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

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