Let OA = a, OB = 10a + 2b and OC = b where O, A and C are non-collinear points. Let p denote the are

Q: Wait, why can't we just use the coordinate geometry formula for area?

While you could technically assign coordinates, it is a trap! It adds unnecessary complexity. Vector algebra allows us to work with the 'soul' of the geometry without needing specific coordinate values. It is faster, cleaner, and less prone to calculation errors.

Q: Why does the cross product of a vector with itself vanish?

Think geometrically! The cross product magnitude represents the area of a parallelogram. If the two sides are identical, the parallelogram collapses into a line segment, which has zero area. Thus, $\vec{a} \times \vec{a} = 0$.

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

Great question! The cross product is anti-commutative, meaning $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. However, since we are taking the magnitude (the absolute value), the negative sign disappears. So, $|\vec{a} \times \vec{b}|$ is the same as $|\vec{b} \times \vec{a}|$. You are safe!

Question

Let $OA = a$, $OB = 10a + 2b$ and $OC = b$ where $O, A$ and $C$ are non-collinear points. Let $p$ denote the area of the quadrilateral $OABC$, and let $q$ denote the area of the parallelogram with $OA$ and $OC$ as adjacent sides. If $p = kq$, then $k = .........$

Sigma Percentile · Accepted Answer

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

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering the hidden elegance of vector geometry.

When you look at a quadrilateral like $OABC$ defined by vectors, it is easy to feel overwhelmed. But remember, in the world of JEE Advanced, complexity is often just a mask for simplicity. Let us peel back that mask together.

### Phase 1: The Foundation

Imagine you are standing at the origin $O$. You have two vectors, $\vec{a}$ and $\vec{b}$, which define your world.

The problem introduces $q$, the area of the parallelogram formed by these vectors. We know from our fundamental toolkit that the area of a parallelogram spanned by $\vec{a}$ and $\vec{b}$ is simply the magnitude of their cross product:

$$q = |\vec{a} 	imes \vec{b}|$$

This $q$ is our reference, our unit of measurement. Everything else will be expressed in terms of this $q$.

### Phase 2: The Divide and Conquer Strategy

We are given a third vector, $\vec{OB} = 10\vec{a} + 2\vec{b}$. This vector defines the point $B$, which completes our quadrilateral $OABC$.

Now, how do we find the area $p$ of this quadrilateral? The secret is to stop looking at it as a single, intimidating shape.

Instead, draw the diagonal $OB$. Suddenly, the quadrilateral splits into two manageable triangles: $	riangle OAB$ and $	riangle OBC$. The total area $p$ is simply the sum of these two:

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

### Phase 3: The Calculation of $	riangle OAB$

Let us focus on $	riangle OAB$. The area of any triangle formed by vectors $\vec{u}$ and $\vec{v}$ is $\frac{1}{2} |\vec{u} 	imes \vec{v}|$.

For our triangle, this becomes $\frac{1}{2} |\vec{OA} 	imes \vec{OB}|$. Substituting our known values, we get:

$$	ext{Area}(	riangle OAB) = \frac{1}{2} |\vec{a} 	imes (10\vec{a} + 2\vec{b})|$$

Here is where the magic happens. Distribute the cross product: $10(\vec{a} 	imes \vec{a}) + 2(\vec{a} 	imes \vec{b})$.

Because $\vec{a} 	imes \vec{a} = 0$, the first term vanishes into thin air! We are left with $\frac{1}{2} |2(\vec{a} 	imes \vec{b})|$, which simplifies beautifully to $|\vec{a} 	imes \vec{b}|$. Since this is $q$, we have found that $	ext{Area}(	riangle OAB) = q$.

### Phase 4: The Calculation of $	riangle OBC$

Now, let us turn to $	riangle OBC$. Its area is $\frac{1}{2} |\vec{OB} 	imes \vec{OC}|$.

Substituting our vectors, we get:

$$	ext{Area}(	riangle OBC) = \frac{1}{2} |(10\vec{a} + 2\vec{b}) 	imes \vec{b}|$$

Again, we distribute the cross product: $10(\vec{a} 	imes \vec{b}) + 2(\vec{b} 	imes \vec{b})$. Just like before, $\vec{b} 	imes \vec{b} = 0$.

We are left with $\frac{1}{2} |10(\vec{a} 	imes \vec{b})|$, which simplifies to $5|\vec{a} 	imes \vec{b}|$. Thus, $	ext{Area}(	riangle OBC) = 5q$.

### The Grand Synthesis

We have arrived at the finish line. The total area $p$ is the sum of our two triangles:

$$p = q + 5q = 6q$$

The problem states $p = kq$, and by simple comparison, we see that **$k = 6$**.

You see? By breaking the problem down and trusting the properties of vectors, we turned a complex geometric puzzle into a simple, elegant result. Keep this mindset, and no problem will ever be too difficult for you.

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors $a$ and $b$

Defining Area $q$ of the Parallelogram

Introducing Vector $O B$ and Quadrilateral $O A B C$

Splitting Quadrilateral $O A B C$ into Triangles

Setting up Area of $△ O A B$

Substituting Vectors for $△ O A B$

Executing Cross Product for $△ O A B$

Finalizing Area of $△ O A B$

Setting up Area of $△ O B C$

Executing Cross Product for $△ O B C$

Finalizing Area of $△ O B C$

Calculating Total Area $p$

Finding the Value of $k$

The Sigma Insight: Vector (Cross) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Synthesis

Similar Questions

Let $O A = a$ , $O B = 12 a + 4 b$ and $O C = b$ , where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $O A$ and $O C$ , then $\frac{area of the quadrilateral OABC}{area of S}$ is equal to

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 :

Let $a = \hat{i} + α \hat{j} + 3 \hat{k}$ and $b = 3 \hat{i} - α \hat{j} + \hat{k}$ . If the area of the parallelogram whose adjacent sides are represented by the vectors $a$ and $b$ is $83$ square units, then $a \cdot b$ is equal to

Let $A (2, 3, 5)$ and $C (- 3, 4, - 2)$ be opposite vertices of a parallelogram $A B C D$ if the diagonal $B D = \hat{i} + 2 \hat{j} + 3 \hat{k}$ then the area of the parallelogram is equal to

Let $O$ be the origin and $O A = 2 \hat{i} + 2 \hat{j} + \hat{k}$ , $O B = \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $O C = \frac{1}{2} (O B - λ O A)$ for some $λ > 0$ . If $∣ O B \times O C ∣ = \frac{9}{2}$ , then which of the following statements is (are) TRUE?

$A (2, 6, 2), B (- 4, 0, λ), C (2, 3, - 1)$ and $D (4, 5, 0), ∣ λ ∣ \leq 5$ are the vertices of a quadrilateral ABCD. If its area is 18 square units, then $5 - 6 λ$ is equal to

Let for a triangle $A B C$ $A B = - 2 \hat{i} + \hat{j} + 3 \hat{k}$ , $C B = α \hat{i} + β \hat{j} + γ \hat{k}$ , $C A = 4 \hat{i} + 3 \hat{j} + δ \hat{k}$ . If $δ > 0$ and the area of the triangle $A B C$ is $56$ then $C B \cdot C A$ is equal to

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that $\frac{Q A}{A R} = \frac{R B}{B P} = \frac{P C}{C Q} = \frac{1}{2}$ . Then $\frac{Area ( Δ P QR )}{Area ( Δ A B C )}$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Defining Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q of the Parallelogram

Introducing Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OB and Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OABC

Splitting Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OABC into Triangles

Setting up Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Substituting Vectors for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Executing Cross Product for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Finalizing Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OAB

Setting up Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OBC

Executing Cross Product for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OBC

Finalizing Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OBC

Calculating Total Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k