If A, B and C are vectors such that |B| = |C|. Prove that [(A + B) x (A + C)] x (B x C) . (B + C) =

Q: Why can't I just expand the whole expression at once? It looks like standard algebra.

That is the classic trap! If you try to expand the entire expression at once, you will end up with a forest of terms that will likely lead to a sign error or a lost vector. In JEE Advanced, the key is 'divide and conquer.' By defining $\vec{V}$, we isolate the complexity, solve it, and then re-integrate it. It is about maintaining clarity in the face of chaos.

Q: I always mess up the order in cross products. Is there a trick?

Yes, the trick is to respect the anti-commutative property: $\vec{B} \times \vec{A} = -\vec{A} \times \vec{B}$. Never assume commutativity. Whenever you swap vectors, pay the 'tax' of a negative sign. If you do this consistently, you will never lose a term.

Q: Why does the term $(\vec{B} \times \vec{C}) \times (\vec{B} \times \vec{C})$ vanish? I thought cross products were non-zero.

A cross product is zero if the two vectors are parallel. A vector is always parallel to itself! Therefore, any vector $\vec{X}$ crossed with itself, $\vec{X} \times \vec{X}$, is always the zero vector. It is a beautiful property that often simplifies the most terrifying-looking expressions.

Question

If $A, B$ and $C$ are vectors such that $|\vec{B}| = |\vec{C}|$. Prove that $[(\vec{A} + \vec{B}) 	imes (\vec{A} + \vec{C})] 	imes (\vec{B} 	imes \vec{C}) \cdot (\vec{B} + \vec{C}) = 0$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system with three vectors: $\vec{A}$, $\vec{B}$, and $\vec{C}$. The problem presents a massive, intimidating expression:

$$[(\vec{A} + \vec{B}) 	imes (\vec{A} + \vec{C})] 	imes (\vec{B} 	imes \vec{C}) \cdot (\vec{B} + \vec{C})$$

Before touching any algebraic rules, look at the constraint: $|\vec{B}| = |\vec{C}|$. This is the heartbeat of the problem, indicating that $\vec{B}$ and $\vec{C}$ are 'twins' in magnitude, living on the same sphere centered at the origin.

### Taming the Beast

We do not attack the monster head-on; we break it down. Let us define an intermediate vector, $\vec{V} = (\vec{A} + \vec{B}) 	imes (\vec{A} + \vec{C})$.

Using the distributive law of cross products, we expand this:

$$\vec{V} = \vec{A} 	imes \vec{A} + \vec{A} 	imes \vec{C} + \vec{B} 	imes \vec{A} + \vec{B} 	imes \vec{C}$$

The term $\vec{A} 	imes \vec{A}$ is zero because the angle between a vector and itself is zero. The expression simplifies to:

$$\vec{V} = \vec{A} 	imes \vec{C} + \vec{B} 	imes \vec{A} + \vec{B} 	imes \vec{C}$$

Using the anti-commutative property to write $\vec{B} 	imes \vec{A}$ as $-\vec{A} 	imes \vec{B}$, our vector $\vec{V}$ becomes:

$$\vec{V} = \vec{A} 	imes \vec{C} - \vec{A} 	imes \vec{B} + \vec{B} 	imes \vec{C}$$

### The Quadruple Product

Now, we return to the main expression and compute $\vec{V} 	imes (\vec{B} 	imes \vec{C})$. Substituting our simplified $\vec{V}$ and distributing the cross product, we generate three separate quadruple products.

The term $(\vec{B} 	imes \vec{C}) 	imes (\vec{B} 	imes \vec{C})$ vanishes instantly. We apply the vector quadruple product identity:

$$(\vec{P} 	imes \vec{Q}) 	imes (\vec{R} 	imes \vec{S}) = [\vec{P} \vec{R} \vec{S}]\vec{Q} - [\vec{Q} \vec{R} \vec{S}]\vec{P}$$

Applying this to the remaining terms, scalar triple products involving repeated vectors (like $[\vec{C} \vec{B} \vec{C}]$ or $[\vec{B} \vec{B} \vec{C}]$) vanish. What remains is a symmetric result:

$$[\vec{A} \vec{B} \vec{C}]\vec{C} - [\vec{A} \vec{B} \vec{C}]\vec{B} = [\vec{A} \vec{B} \vec{C}](\vec{C} - \vec{B})$$

### The Grand Finale

We take our result and dot it with $(\vec{B} + \vec{C})$. The expression becomes:

$$[\vec{A} \vec{B} \vec{C}](\vec{C} - \vec{B}) \cdot (\vec{C} + \vec{B})$$

Focusing on the dot product part, $(\vec{C} - \vec{B}) \cdot (\vec{C} + \vec{B})$, we expand it as:

$$\vec{C} \cdot \vec{C} + \vec{C} \cdot \vec{B} - \vec{B} \cdot \vec{C} - \vec{B} \cdot \vec{B} = |\vec{C}|^2 - |\vec{B}|^2$$

Because the problem stated $|\vec{B}| = |\vec{C}|$, this term is zero. Consequently, the entire expression collapses to **0**.

Animated Solution for Mathematics - Vector Algebra: If $A, B$ and $C$ are vectors such that $∣ B ∣ = ∣ C ∣$ . Prove that $[(A + B) \times (A + C)] \times (B \times C) \cdot (B + C) = 0$ .

Visualized Solution

Visualizing the Vectors

Defining Intermediate Vector $V$

Expanding $V$

Simplifying $A \times A$

Rearranging Cross Products

Setting up the Next Cross Product

Distributing the Cross Product

Eliminating the Third Term

Vector Quadruple Product Identity

Evaluating the First Term

Evaluating the Second Term

Combining the Terms

The Final Dot Product

Expanding the Dot Product

Applying the Magnitude Condition

The Sigma Insight: Vector Triple Product

Analyzing the Setup

Taming the Beast

The Quadruple Product

The Grand Finale

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