If u, v and w are three non-coplanar vectors, then (u + v - w) . [(u - v) x (v - w)] equals (a) 3u.v

Q: Wait, why do we reject the terms with repeated vectors? Is there a faster way?

That is a brilliant observation! Yes, you can think of it geometrically. The Scalar Triple Product $\vec{a} \cdot (\vec{b} \times \vec{c})$ represents the volume of a parallelepiped formed by $\vec{a}, \vec{b},$ and $\vec{c}$. If any two vectors are the same, the parallelepiped collapses into a flat shape with zero volume. Algebraically, $\vec{b} \times \vec{c}$ is always perpendicular to both $\vec{b}$ and $\vec{c}$. If $\vec{a}$ is one of those vectors, the dot product becomes $\vec{a} \cdot (\text{something perpendicular to } \vec{a})$, which is always zero. It is not just a trick; it is a fundamental geometric truth!

Q: Does the order of vectors in the cross product really matter that much?

It matters immensely! The cross product is anti-commutative, meaning $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$. If you flip the order, you flip the sign. When expanding $(\vec{u} - \vec{v}) \times (\vec{v} - \vec{w})$, you must treat it like a binomial expansion but respect the non-commutative nature of the cross product. Always keep the left-to-right order of the vectors as they appear in the expansion.

Q: How do I know when to use cyclic permutation versus swapping in the box product?

Think of the box product $[\vec{u} \vec{v} \vec{w}]$ as a cycle. A cyclic permutation (shifting $\vec{u} \to \vec{v} \to \vec{w} \to \vec{u}$) preserves the value and the sign. However, swapping any two adjacent vectors (like $\vec{u}$ and $\vec{v}$) reverses the orientation, which introduces a negative sign. If you are ever unsure, just write it out in dot-cross notation—it never fails!

Question

If $\vec{u}, \vec{v}$ and $\vec{w}$ are three non-coplanar vectors, then $(\vec{u} + \vec{v} - \vec{w}) \cdot [(\vec{u} - \vec{v}) 	imes (\vec{v} - \vec{w})]$ equals

Options:

(A) $3\vec{u} \cdot \vec{v} 	imes \vec{w}$

(B) 0

(C) $\vec{u} \cdot \vec{v} 	imes \vec{w}$

(D) $\vec{u} \cdot \vec{w} 	imes \vec{v}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast 3D space, holding three non-coplanar vectors: $\vec{u}, \vec{v},$ and $\vec{w}$. Because they are non-coplanar, they do not lie flat on a single sheet of paper; instead, they reach out into the third dimension, defining the edges of a parallelepiped.

Our mission is to evaluate the expression:
$$(\vec{u} + \vec{v} - \vec{w}) \cdot [(\vec{u} - \vec{v}) 	imes (\vec{v} - \vec{w})]$$

It looks intimidating, but like all great challenges in JEE mathematics, the secret lies in breaking it down systematically.

### Phase 1: The Cross Product Expansion

Let us focus on the heart of the expression: the cross product inside the square brackets, $(\vec{u} - \vec{v}) 	imes (\vec{v} - \vec{w})$. Just as you would expand $(a-b)(c-d)$ in algebra, we distribute the cross product.

However, we must be vigilant. The cross product is not commutative, so the order of vectors is sacred. Expanding term by term, we get:
$$\vec{u} 	imes \vec{v} - \vec{u} 	imes \vec{w} - \vec{v} 	imes \vec{v} + \vec{v} 	imes \vec{w}$$

Look closely at the third term, $\vec{v} 	imes \vec{v}$. Since the angle between a vector and itself is zero, and $\sin(0) = 0$, this term vanishes into the zero vector $\vec{0}$. Our expression simplifies beautifully to:
$$\vec{u} 	imes \vec{v} - \vec{u} 	imes \vec{w} + \vec{v} 	imes \vec{w}$$

### Phase 2: The Dot Product and the Zero-Volume Insight

Now, we bring back the first part of our expression:
$$(\vec{u} + \vec{v} - \vec{w}) \cdot (\vec{u} 	imes \vec{v} - \vec{u} 	imes \vec{w} + \vec{v} 	imes \vec{w})$$

If we were to expand this fully, we would have nine terms. But we are smarter than that. We know the property of the Scalar Triple Product (STP): any term containing a repeated vector is zero because the volume of the resulting parallelepiped is zero.

For example, $\vec{u} \cdot (\vec{u} 	imes \vec{v}) = 0$. By filtering out these zero terms, we are left with only the combinations where all three vectors are distinct:
$$\vec{u} \cdot (\vec{v} 	imes \vec{w}) - \vec{v} \cdot (\vec{u} 	imes \vec{w}) - \vec{w} \cdot (\vec{u} 	imes \vec{v})$$

### Phase 3: The Elegance of Box Notation

Let us translate this into the elegant box notation:
$$[\vec{u} \vec{v} \vec{w}] - [\vec{v} \vec{u} \vec{w}] - [\vec{w} \vec{u} \vec{v}]$$

To combine these, we use the properties of the STP. Swapping two vectors changes the sign, and cyclic permutation preserves it. The second term, $-[\vec{v} \vec{u} \vec{w}]$, becomes $+[\vec{u} \vec{v} \vec{w}]$ after swapping $\vec{u}$ and $\vec{v}$.

The third term, $-[\vec{w} \vec{u} \vec{v}]$, is a cyclic permutation of $-[\vec{u} \vec{v} \vec{w}]$. Substituting these back, we have:
$$[\vec{u} \vec{v} \vec{w}] + [\vec{u} \vec{v} \vec{w}] - [\vec{u} \vec{v} \vec{w}]$$

The result is simply $[\vec{u} \vec{v} \vec{w}]$, or **$\vec{u} \cdot (\vec{v} 	imes \vec{w})$**. We have successfully navigated the complexity and arrived at a clean, elegant solution.

Animated Solution for Mathematics - Vector Algebra: If $u, v$ and $w$ are three non-coplanar vectors, then $(u + v - w) \cdot [(u - v) \times (v - w)]$ equals

Select Answer:

Visualized Solution

Visualizing the Vectors

Expanding the Cross Product

Distributing the Terms

Eliminating the Zero Vector

The Simplified Cross Product

Setting up the Dot Product

The Zero Property of STP

Extracting Non-Zero Terms

Converting to Box Notation

Swapping Vectors in STP

Cyclic Permutation in STP

Final Calculation

Conclusion

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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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 u,v and w are three non-coplanar vectors, then (u+v−w)⋅[(u−v)×(v−w)] equals