Let a, b, c be three coplanar concurrent vectors such that angles between any two of them is same. I

Q: Wait, why is the angle between the vectors $120^\circ$? Shouldn't it be $60^\circ$?

Great question! Remember, the vectors are concurrent and coplanar. If you place them on a 2D plane originating from a single point, they divide the full $360^\circ$ rotation into three sectors. Since the problem states the angle between *any* two is the same, we divide $360^\circ$ by 3, which gives $120^\circ$. If you chose $60^\circ$, you would be looking at the angle between the vectors and their extensions, not the vectors themselves!

Q: Why can't I just calculate the cross products $\vec{a} \times \vec{b}$ directly?

You could, but it would be a nightmare of trigonometric components! By using Lagrange's Identity, we bypass the need to define a coordinate system or deal with sines and cosines of components. We transform a complex vector operation into a simple algebraic expression involving only magnitudes and dot products. Always look for the identity that simplifies the geometry!

Q: How do I know the cyclic symmetry holds for the other two terms?

Because the problem is perfectly symmetric! The relationship between $\vec{a}, \vec{b},$ and $\vec{c}$ is identical for every pair. If the first term $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c})$ yields $\frac{3}{4}ab^2c$, then by simply rotating the labels, the second term $(\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a})$ must yield $\frac{3}{4}bc^2a$. Trust the symmetry; it is the hallmark of an elegant solution.

Question

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} 	imes \vec{b}) \cdot (\vec{b} 	imes \vec{c}) + (\vec{b} 	imes \vec{c}) \cdot (\vec{c} 	imes \vec{a}) + (\vec{c} 	imes \vec{a}) \cdot (\vec{a} 	imes \vec{b}) = 168$, then $|\vec{a}| + |\vec{b}| + |\vec{c}|$ is equal to:

Options:

(A) 10

(B) 14

(C) 16

(D) 18

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a 2D plane. Three vectors, $\vec{a}, \vec{b},$ and $\vec{c}$, emerge from your feet like spokes on a wheel. They are coplanar and concurrent, sharing the same starting point.

The problem states that the angle between any two of these vectors is identical. Since they span a full $360^\circ$ circle, we divide that circle by three. Each pair of vectors is separated by exactly $120^\circ$.

### The Master Key: Lagrange's Identity

We are confronted with the expression:
$$(\vec{a} 	imes \vec{b}) \cdot (\vec{b} 	imes \vec{c}) + (\vec{b} 	imes \vec{c}) \cdot (\vec{c} 	imes \vec{a}) + (\vec{c} 	imes \vec{a}) \cdot (\vec{a} 	imes \vec{b}) = 168$$

We utilize Lagrange's Identity, which states that the dot product of two cross products can be rewritten as a determinant of dot products:
$$(\vec{p} 	imes \vec{q}) \cdot (\vec{r} 	imes \vec{s}) = (\vec{p} \cdot \vec{r})(\vec{q} \cdot \vec{s}) - (\vec{p} \cdot \vec{s})(\vec{q} \cdot \vec{r})$$

This identity acts as the bridge between the complex world of cross products and the simpler world of magnitudes and dot products.

### The Expansion

Let us focus on the first term: $(\vec{a} 	imes \vec{b}) \cdot (\vec{b} 	imes \vec{c})$. Using the identity with $\vec{p}=\vec{a}, \vec{q}=\vec{b}, \vec{r}=\vec{b},$ and $\vec{s}=\vec{c}$, we obtain:
$$(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b})$$

We know the dot product $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos(120^\circ) = -\frac{1}{2}|\vec{u}||\vec{v}|$. Thus, $\vec{a} \cdot \vec{b} = -\frac{1}{2}ab$, $\vec{b} \cdot \vec{c} = -\frac{1}{2}bc$, and $\vec{a} \cdot \vec{c} = -\frac{1}{2}ac$. Also, $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = b^2$.

Substituting these values, the expression becomes:
$$\left(-\frac{1}{2}abight)\left(-\frac{1}{2}bcight) - \left(-\frac{1}{2}acight)(b^2) = \frac{1}{4}ab^2c + \frac{1}{2}ab^2c = \frac{3}{4}ab^2c$$

### The Power of Symmetry

Because the setup is perfectly symmetrical, we do not need to repeat this process for the other two terms. By cyclic symmetry, the second term is $\frac{3}{4}bc^2a$ and the third term is $\frac{3}{4}ca^2b$.

Adding them all together, we get:
$$\frac{3}{4}ab^2c + \frac{3}{4}bc^2a + \frac{3}{4}ca^2b$$

Factoring out $\frac{3}{4}abc$ from this sum, we are left with:
$$\frac{3}{4}abc(a + b + c)$$

### The Final Victory

We are given that the product of the magnitudes $abc = 14$ and the total sum equals $168$. Substituting these into our simplified expression:
$$\frac{3}{4}(14)(a + b + c) = 168$$

Calculating $\frac{3}{4} 	imes 14$, we get $10.5$. Thus:
$$10.5(a + b + c) = 168$$

Dividing $168$ by $10.5$, we arrive at our final answer:
**$a + b + c = 16$**

Select Answer:

Visualized Solution

Visualizing the Vector Configuration

Defining Magnitudes and Dot Products

Dot Products for All Pairs

Introducing Lagrange's Identity

Applying Identity to the First Term

Substituting Dot Products

Simplifying the First Term

Summing All Terms by Symmetry

Factoring the Total Sum

Final Substitution and Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Master Key

The Expansion

The Power of Symmetry

The Final Victory

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