If A, B, C, D are any four points in space, prove that |AB × CD + BC × AD + CA × BD| = 4 (area of tr

Q: Wait, why does $\vec{d}$ disappear? Does this mean the position of $D$ doesn't matter?

That is a brilliant observation! Yes, the result is independent of the position of $D$. Geometrically, this makes sense because the expression represents a property of the triangle $ABC$ itself. Since $D$ is an arbitrary point, its contribution to the cross-product sum must net out to zero. It is a beautiful demonstration of how algebraic manipulation reveals geometric truth.

Q: Why can't we just use the geometric definition of the cross product directly?

You could, but it would be a nightmare! Trying to visualize the cross product of vectors in 3D space without an algebraic framework is like trying to solve a maze blindfolded. The algebraic method (using position vectors) is the 'superpower' that allows us to handle the geometry without needing to visualize complex 3D orientations.

Q: I'm stuck on the area formula. Why is the area of $\Delta ABC$ equal to $\frac{1}{2} |\vec{b} \times \vec{a} + \vec{c} \times \vec{b} + \vec{a} \times \vec{c}|$?

Great question. The standard formula is $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. If you expand $\vec{AB} \times \vec{AC}$ using position vectors $(\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})$, you get $\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a}$. Since $\vec{a} \times \vec{a} = 0$, and using the anti-commutative property, this simplifies exactly to the cyclic sum form. It's a standard identity in vector geometry!

Question

If $A, B, C, D$ are any four points in space, prove that $|\vec{AB} 	imes \vec{CD} + \vec{BC} 	imes \vec{AD} + \vec{CA} 	imes \vec{BD}| = 4 (	ext{area of triangle } ABC)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving a problem; we are embarking on a journey through the elegant architecture of vector space. When you look at the expression $|\vec{AB} 	imes \vec{CD} + \vec{BC} 	imes \vec{AD} + \vec{CA} 	imes \vec{BD}|$, it is natural to feel a sense of unease.

In the world of JEE Advanced, complexity is often just a mask for a hidden, beautiful simplicity. Let us peel back that mask together.

### The Power of Position Vectors

Imagine four points, $A, B, C,$ and $D$, floating in the vast expanse of 3D space. To tame this chaos, we assign position vectors $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ to these points.

By anchoring these points to an origin, we translate the geometric concept of a segment into the algebraic language of subtraction. We define our building blocks as follows:

$$\vec{AB} = \vec{b} - \vec{a}, \quad \vec{CD} = \vec{d} - \vec{c}$$
$$\vec{BC} = \vec{c} - \vec{b}, \quad \vec{AD} = \vec{d} - \vec{a}$$
$$\vec{CA} = \vec{a} - \vec{c}, \quad \vec{BD} = \vec{d} - \vec{b}$$

With these definitions, we have successfully converted the geometric problem into an algebraic expression. We are no longer looking at lines in space; we are looking at a system of equations waiting to be solved.

### The Algebraic Assault

Now, we substitute these definitions into our main expression:

$$|(\vec{b}-\vec{a}) 	imes (\vec{d}-\vec{c}) + (\vec{c}-\vec{b}) 	imes (\vec{d}-\vec{a}) + (\vec{a}-\vec{c}) 	imes (\vec{d}-\vec{b})|$$

Treat this like a standard algebraic expansion using the distributive property: $(\vec{u} - \vec{v}) 	imes (\vec{w} - \vec{z}) = \vec{u} 	imes \vec{w} - \vec{u} 	imes \vec{z} - \vec{v} 	imes \vec{w} + \vec{v} 	imes \vec{z}$.

Expanding the first term, we get:
$$(\vec{b}-\vec{a}) 	imes (\vec{d}-\vec{c}) = \vec{b} 	imes \vec{d} - \vec{b} 	imes \vec{c} - \vec{a} 	imes \vec{d} + \vec{a} 	imes \vec{c}$$

Repeating this for the second and third terms:
$$(\vec{c}-\vec{b}) 	imes (\vec{d}-\vec{a}) = \vec{c} 	imes \vec{d} - \vec{c} 	imes \vec{a} - \vec{b} 	imes \vec{d} + \vec{b} 	imes \vec{a}$$
$$(\vec{a}-\vec{c}) 	imes (\vec{d}-\vec{b}) = \vec{a} 	imes \vec{d} - \vec{a} 	imes \vec{b} - \vec{c} 	imes \vec{d} + \vec{c} 	imes \vec{b}$$

### The Vanishing Act

Look at the sum of all these expanded terms. Notice that the terms involving $\vec{d}$ (such as $\vec{b} 	imes \vec{d}$ and $-\vec{b} 	imes \vec{d}$) cancel out entirely.

The variable $\vec{d}$—the point that seemed to define the complexity of the problem—vanishes. This tells us that the result is independent of the position of $D$.

After simplifying using the property $\vec{x} 	imes \vec{y} = -\vec{y} 	imes \vec{x}$, our expression collapses into:

$$2|\vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{b} + \vec{a} 	imes \vec{c}|$$

### The Geometric Bridge

We connect this result to the area of triangle $ABC$. Recall the fundamental vector formula for the area of a triangle:

$$	ext{Area} = \frac{1}{2} |\vec{BC} 	imes \vec{BA}|$$

Substituting our position vectors, we get:
$$	ext{Area} = \frac{1}{2} |(\vec{c}-\vec{b}) 	imes (\vec{a}-\vec{b})| = \frac{1}{2} |\vec{c} 	imes \vec{a} - \vec{c} 	imes \vec{b} - \vec{b} 	imes \vec{a} + \vec{b} 	imes \vec{b}|$$

Since $\vec{b} 	imes \vec{b} = 0$, this simplifies to:
$$	ext{Area} = \frac{1}{2} |\vec{c} 	imes \vec{a} + \vec{b} 	imes \vec{c} + \vec{a} 	imes \vec{b}|$$

### Final Calculation

We have shown that our original sum is $2 	imes |\vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{b} + \vec{a} 	imes \vec{c}|$. Simultaneously, the area of the triangle is $\frac{1}{2} |\vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{b} + \vec{a} 	imes \vec{c}|$.

Therefore, the original sum is equal to **$4 	imes 	ext{Area}(\Delta ABC)$**.

This is the essence of mathematics: trusting the process, staying calm under pressure, and watching as the pieces of the puzzle fall perfectly into place. You have mastered this.

Animated Solution for Mathematics - Vector Algebra: If $A, B, C, D$ are any four points in space, prove that $∣ A B \times C D + B C \times A D + C A \times B D ∣ = 4 (area of triangle A B C)$ .

Visualized Solution

Defining Position Vectors

Expressing Segments in Vector Form

Expressing More Segments

Final Segment Pairs

Substituting into the Expression

Expanding the First Term

Expanding the Second Term

Expanding the Third Term

Summation and Cancellation

Simplifying the Remaining Terms

Using Cross Product Properties

Area of Triangle $A B C$

Expanding the Area Formula

Simplifying the Area Expression

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of Position Vectors

The Algebraic Assault

The Vanishing Act

The Geometric Bridge

Final Calculation

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A,B,C,D are any four points in space, prove that ∣AB×CD+BC×AD+CA×BD∣=4(area of triangle ABC).