Let a, b, c ∈ R be such that a² + b² + c² = 1. If a cos θ = b cos (θ + 2π/3) = c cos (θ + 4π/3), whe

Q: Wait, why do we reject the negative root here?

Actually, we don't reject anything! The magnitude is always positive by definition, $\sqrt{a^2+b^2+c^2}$. Since $a^2+b^2+c^2=1$, the magnitude is simply $1$. No roots to reject, just a beautiful simplification.

Q: Why can't we just use energy conservation?

Energy conservation is for physical systems involving forces and work. Here, we are dealing with pure vector geometry and trigonometry. The 'energy' of this problem lies in the dot product, which measures the alignment of these vectors.

Q: Is there a faster way than the long trigonometric expansion?

You could use complex numbers or phasors, representing the terms as real parts of $e^{i\theta}$. However, the trigonometric identity approach is the most direct and elegant way to see the cancellation happen.

Question

Let $a, b, c \in R$ be such that $a^2 + b^2 + c^2 = 1$. If $a \cos 	heta = b \cos (	heta + \frac{2\pi}{3}) = c \cos (	heta + \frac{4\pi}{3})$, where $	heta = \frac{\pi}{9}$, then the angle between the vectors $a\hat{i} + b\hat{j} + c\hat{k}$ and $b\hat{i} + c\hat{j} + a\hat{k}$ is :

Options:

(A) 0

(B) $\frac{\pi}{9}$

(C) $\frac{\pi}{2}$

(D) $\frac{2\pi}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space, looking at two vectors, $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$.

At first glance, they seem like arbitrary collections of variables. But look closer. The problem gives us a powerful anchor: $a^2 + b^2 + c^2 = 1$.

This is not just an equation; it is a geometric statement. It tells us that the magnitude of both $\vec{u}$ and $\vec{v}$ is exactly $1$. They are unit vectors!

This realization is our first step toward victory. We are not dealing with complex, scaling vectors; we are dealing with vectors that live on the surface of a unit sphere.

### The Dot Product Bridge

We want to find the angle $\phi$ between these two vectors. The most reliable tool in our arsenal is the dot product formula:

$$\cos \phi = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$$

Since we have already established that $|\vec{u}| = 1$ and $|\vec{v}| = 1$, the denominator vanishes, leaving us with the elegant result: $\cos \phi = \vec{u} \cdot \vec{v}$.

Calculating the dot product is straightforward: we multiply the corresponding components and sum them up. Thus, $\cos \phi = ab + bc + ca$.

Now, the problem reduces to finding the value of $ab + bc + ca$. This is where the trigonometric constraint comes into play.

### The Algebraic Symmetry

We are given the constraint:

$$a \cos 	heta = b \cos \left( 	heta + \frac{2\pi}{3} ight) = c \cos \left( 	heta + \frac{4\pi}{3} ight)$$

This looks intimidating, but let's use a classic algebraic trick: set the entire expression equal to a constant $k$. This allows us to isolate $a, b,$ and $c$ as:

$$a = \frac{k}{\cos 	heta}, \quad b = \frac{k}{\cos \left( 	heta + \frac{2\pi}{3} ight)}, \quad c = \frac{k}{\cos \left( 	heta + \frac{4\pi}{3} ight)}$$

Now, substitute these into our expression for $\cos \phi$. We get:

$$\cos \phi = k^2 \left[ \frac{1}{\cos 	heta \cos \left( 	heta + \frac{2\pi}{3} ight)} + \frac{1}{\cos \left( 	heta + \frac{2\pi}{3} ight) \cos \left( 	heta + \frac{4\pi}{3} ight)} + \frac{1}{\cos \left( 	heta + \frac{4\pi}{3} ight) \cos 	heta} ight]$$

It looks messy, but don't panic!

### The Beautiful Cancellation

When we take the common denominator, the numerator becomes:

$$\cos \left( 	heta + \frac{4\pi}{3} ight) + \cos 	heta + \cos \left( 	heta + \frac{2\pi}{3} ight)$$

This is a classic sum of cosines with angles in an arithmetic progression. Using the identity $\cos C + \cos D = 2 \cos \left( \frac{C+D}{2} ight) \cos \left( \frac{C-D}{2} ight)$, we can simplify the sum of the last two terms.

The sum becomes $-\cos 	heta$. When we add this to the first term, $\cos 	heta$, the result is exactly zero!

The entire numerator vanishes. This means $\cos \phi = 0$, which implies that **$\phi = \frac{\pi}{2}$**.

The vectors are perfectly perpendicular. A complex algebraic setup has collapsed into a simple, elegant geometric truth.

Select Answer:

Visualized Solution

Defining the Vectors

Magnitudes of $u$ and $v$

The Dot Product Formula

Calculating the Dot Product

The Trigonometric Constraint

Isolating $a, b,$ and $c$

Substituting into $cos ϕ$

Taking the Common Denominator

Focusing on the Numerator

Applying Trigonometric Identity

Simplifying the Terms

The Numerator Vanishes

The Final Angle

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Dot Product Bridge

The Algebraic Symmetry

The Beautiful Cancellation

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