If the vectors b, c, d are not coplanar, then prove that the vector (a x b) x (c x d) + (a x c) x (d

Q: Wait, why do we reject the negative root here? Or rather, why does the expression simplify to a multiple of \vec{a}?

That is a fantastic question! We aren't rejecting anything; we are simplifying. The expression doesn't just 'happen' to be parallel to \vec{a}; it is mathematically forced to be. When we use the Basis Expansion Theorem, we express the sum as a scalar multiple of \vec{a}. Since any vector \vec{v} = k\vec{a} is by definition parallel to \vec{a}, the proof is complete. The scalar here is -2[\vec{b} \vec{c} \vec{d}].

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm! However, this is a pure vector algebra problem. Physics principles like energy conservation apply to systems with forces and motion. Here, we are dealing with the geometric properties of space itself. We must rely on the axioms of vector algebra—specifically the quadruple product identity and the basis expansion theorem. Stick to the math, and the geometry will reveal itself!

Q: Is the scalar triple product [\vec{b} \vec{c} \vec{d}] always non-zero?

Yes, absolutely! The problem explicitly states that \vec{b}, \vec{c}, and \vec{d} are non-coplanar. If they were coplanar, they would lie on the same plane, and their volume (which is what the scalar triple product represents) would be zero. Because they are non-coplanar, they span 3D space, and [\vec{b} \vec{c} \vec{d}] \neq 0. This is the crucial condition that allows us to use them as a basis.

Question

If the vectors $\vec{b}, \vec{c}, \vec{d}$ are not coplanar, then prove that the vector $(\vec{a} 	imes \vec{b}) 	imes (\vec{c} 	imes \vec{d}) + (\vec{a} 	imes \vec{c}) 	imes (\vec{d} 	imes \vec{b}) + (\vec{a} 	imes \vec{d}) 	imes (\vec{b} 	imes \vec{c})$ is parallel to $\vec{a}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the realm of JEE Advanced mathematics. Today, we are not just solving a problem; we are conducting a symphony of vectors.

When you first look at the expression 
$$(\vec{a} 	imes \vec{b}) 	imes (\vec{c} 	imes \vec{d}) + (\vec{a} 	imes \vec{c}) 	imes (\vec{d} 	imes \vec{b}) + (\vec{a} 	imes \vec{d}) 	imes (\vec{b} 	imes \vec{c})$$ 
it looks like a chaotic mess of cross products. It is intimidating and designed to make you doubt your intuition.

However, in the world of JEE, complexity is often just a mask for elegance. Let us peel back that mask.

### The Weapon of Choice

To dismantle this beast, we need a scalpel, not a sledgehammer. That scalpel is the **Vector Quadruple Product Identity**.

The identity states that for any four vectors:
$$(\vec{x} 	imes \vec{y}) 	imes (\vec{z} 	imes \vec{w}) = [\vec{x} \vec{z} \vec{w}]\vec{y} - [\vec{y} \vec{z} \vec{w}]\vec{x}$$

Look at what this does! It takes a terrifying cross product of cross products and transforms it into a simple linear combination of the first two vectors, $\vec{x}$ and $\vec{y}$. It is the bridge between the complex and the simple.

### The Expansion

Let us take the first term: $(\vec{a} 	imes \vec{b}) 	imes (\vec{c} 	imes \vec{d})$. Using our identity, we treat $\vec{a}$ as $\vec{x}$, $\vec{b}$ as $\vec{y}$, $\vec{c}$ as $\vec{z}$, and $\vec{d}$ as $\vec{w}$.

The expansion becomes:
$$[\vec{a} \vec{c} \vec{d}]\vec{b} - [\vec{b} \vec{c} \vec{d}]\vec{a}$$

Now, let us repeat this for the second term: $(\vec{a} 	imes \vec{c}) 	imes (\vec{d} 	imes \vec{b})$. Here, our $\vec{x}$ is $\vec{a}$, $\vec{y}$ is $\vec{c}$, $\vec{z}$ is $\vec{d}$, and $\vec{w}$ is $\vec{b}$.

The expansion yields:
$$[\vec{a} \vec{d} \vec{b}]\vec{c} - [\vec{c} \vec{d} \vec{b}]\vec{a}$$

Finally, for the third term: $(\vec{a} 	imes \vec{d}) 	imes (\vec{b} 	imes \vec{c})$. Following the same pattern, we get:
$$[\vec{a} \vec{b} \vec{c}]\vec{d} - [\vec{d} \vec{b} \vec{c}]\vec{a}$$

### The Cyclic Symmetry

Now, let us sum these pieces. When we group the terms, we see two distinct parts.

The first part involves the vectors $\vec{b}$, $\vec{c}$, and $\vec{d}$ scaled by various box products:
$$[\vec{a} \vec{c} \vec{d}]\vec{b} + [\vec{a} \vec{d} \vec{b}]\vec{c} + [\vec{a} \vec{b} \vec{c}]\vec{d}$$

The second part involves the vector $\vec{a}$ multiplied by the sum of the box products:
$$-([\vec{b} \vec{c} \vec{d}] + [\vec{c} \vec{d} \vec{b}] + [\vec{d} \vec{b} \vec{c}])\vec{a}$$

Here is where the beauty of the scalar triple product shines. Recall the cyclic property: $[\vec{b} \vec{c} \vec{d}] = [\vec{c} \vec{d} \vec{b}] = [\vec{d} \vec{b} \vec{c}]$. They are all the same, so the second part simplifies beautifully to $-3[\vec{b} \vec{c} \vec{d}]\vec{a}$.

### The Grand Finale

We are almost there. We have:
$$[\vec{a} \vec{c} \vec{d}]\vec{b} + [\vec{a} \vec{d} \vec{b}]\vec{c} + [\vec{a} \vec{b} \vec{c}]\vec{d} - 3[\vec{b} \vec{c} \vec{d}]\vec{a}$$

Does the first part look familiar? It is the **Basis Expansion Theorem**. Since $\vec{b}$, $\vec{c}$, and $\vec{d}$ are non-coplanar, they form a basis, and any vector $\vec{a}$ can be written as:
$$[\vec{b} \vec{c} \vec{d}]\vec{a} = [\vec{a} \vec{c} \vec{d}]\vec{b} + [\vec{a} \vec{d} \vec{b}]\vec{c} + [\vec{a} \vec{b} \vec{c}]\vec{d}$$

Substituting this into our sum, we get:
$$[\vec{b} \vec{c} \vec{d}]\vec{a} - 3[\vec{b} \vec{c} \vec{d}]\vec{a} = -2[\vec{b} \vec{c} \vec{d}]\vec{a}$$

The result is a scalar multiple of $\vec{a}$. We have conquered the monster and found the order within the chaos. The final answer is **$-2[\vec{b} \vec{c} \vec{d}]\vec{a}$**.

Animated Solution for Mathematics - Vector Algebra: If the vectors $b, c, d$ are not coplanar, then prove that the vector $(a \times b) \times (c \times d) + (a \times c) \times (d \times b) + (a \times d) \times (b \times c)$ is parallel to $a$ .

Visualized Solution

Visualizing the Vectors

The Vector Quadruple Product Identity

Expanding $(a \times b) \times (c \times d)$

Expanding $(a \times c) \times (d \times b)$

Expanding $(a \times d) \times (b \times c)$

Summing the Results

Cyclic Property of STP

The Basis Expansion Identity

Final Conclusion

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Weapon of Choice

The Expansion

The Cyclic Symmetry

The Grand Finale

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