Let a, b, c be unit vectors such that a + b + c = 0. Which one of the following is correct?
(a) a ×

Q: Wait, why can't we just say the cross product is zero because the sum is zero?

That is a common trap! The cross product of a sum is the sum of the cross products, not the cross product of the sum. When you expand $\vec{a} \times (\vec{a} + \vec{b} + \vec{c})$, you get $\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$. While the sum is zero, the individual cross products do not necessarily vanish. We must use the properties of vectors to see how they relate to each other.

Q: Why is the angle $120^{\circ}$ and not $60^{\circ}$?

This is a classic geometric pitfall. In the triangle, the vectors are placed head-to-tail, so the internal angle is $60^{\circ}$. However, the definition of the cross product $\vec{a} \times \vec{b}$ requires the vectors to be placed tail-to-tail. When you shift vector $\vec{a}$ to be tail-to-tail with $\vec{b}$, the angle between them becomes the exterior angle, which is $180^{\circ} - 60^{\circ} = 120^{\circ}$.

Q: How do we know the cross products are equal to each other and not just parallel?

The algebra proves they are identical. By crossing the equation with $\vec{a}$, we found $\vec{a} \times \vec{b} = \vec{c} \times \vec{a}$. By crossing with $\vec{b}$, we find $\vec{b} \times \vec{c} = \vec{a} \times \vec{b}$. Since they all equal the same vector, they must be equal in both magnitude and direction, not just parallel.

Question

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. Which one of the following is correct?

Options:

(A) $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} = \vec{0}$

(B) $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} 
eq \vec{0}$

(C) $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{a} 	imes \vec{c} 
eq \vec{0}$

(D) $\vec{a} 	imes \vec{b}, \vec{b} 	imes \vec{c}, \vec{c} 	imes \vec{a}$ are mutually perpendicular

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a beautiful problem. Often in JEE Advanced, we are presented with equations that look simple, yet they hide a profound geometric truth.

Today, we look at three unit vectors, $\vec{a}, \vec{b}, \vec{c}$, whose sum is the zero vector:
$$\vec{a} + \vec{b} + \vec{c} = \vec{0}$$

Imagine these vectors as three forces acting on a point, or perhaps three steps you take in a field. If you take these steps and end up exactly where you started, you have traced a closed loop.

Because these are unit vectors, each step has a length of exactly one. A closed loop of three equal-length vectors is not just any shape; it is a perfect equilateral triangle. This is our anchor.

### The Surgical Strike: The Cross Product

Now, we need to find the relationship between their cross products. How do we extract a cross product from a simple addition equation?

We use a surgical strike. We operate on the entire equation with one of the vectors. Let us choose $\vec{a}$.

We take the cross product of $\vec{a}$ with both sides of our equation:
$$\vec{a} 	imes (\vec{a} + \vec{b} + \vec{c}) = \vec{a} 	imes \vec{0}$$

This is a powerful technique. We are not calculating yet; we are setting the stage. By distributing the cross product, we get:
$$\vec{a} 	imes \vec{a} + \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{c} = \vec{0}$$

### The Elegance of Properties

Here is where the magic happens. We know that any vector crossed with itself is the zero vector, because the angle between them is zero, and $\sin(0^{\circ}) = 0$.

So, $\vec{a} 	imes \vec{a} = \vec{0}$. Our equation simplifies to:
$$\vec{0} + \vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{c} = \vec{0}$$

This leaves us with $\vec{a} 	imes \vec{b} = -(\vec{a} 	imes \vec{c})$. Now, recall the anti-commutative property: the cross product flips sign if you swap the order of the vectors.

Thus, $-(\vec{a} 	imes \vec{c})$ is exactly $\vec{c} 	imes \vec{a}$. We have arrived at:
$$\vec{a} 	imes \vec{b} = \vec{c} 	imes \vec{a}$$

### The Power of Symmetry

If we repeat this process by crossing the original equation with $\vec{b}$, we would find that $\vec{b} 	imes \vec{c} = \vec{a} 	imes \vec{b}$. By the transitive property of equality, we conclude that:
$$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$$

All three cross products are identical. But are they zero? We must check the magnitude.

The magnitude of $\vec{a} 	imes \vec{b}$ is $|\vec{a}||\vec{b}|\sin(	heta)$. As we discussed, the angle $	heta$ between the vectors when placed tail-to-tail is $120^{\circ}$.

Since $\sin(120^{\circ}) = \frac{\sqrt{3}}{2}$, the magnitude is $\frac{\sqrt{3}}{2}$, which is clearly not zero. We have solved it!

**The cross products are equal and non-zero.** This is the elegance of vector algebra—a few simple properties reveal a deep, symmetrical truth.

Animated Solution for Mathematics - Vector Algebra: Let $a, b, c$ be unit vectors such that $a + b + c = 0$ . Which one of the following is correct?

Select Answer:

Visualized Solution

The Vector Triangle

The Cross Product Strategy

Operating with $a$

Distributing the Product

Self Cross Product Property

Simplifying the Equation

Transposing the Term

Anti-commutativity Property

Applying Anti-commutativity

Symmetry for $b$

Combining the Equalities

Checking the Non-Zero Condition

Calculating the Magnitude

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Surgical Strike

The Elegance of Properties

The Power of Symmetry

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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; } Let a,b,c be unit vectors such that a+b+c=0. Which one of the following is correct?