Consider three vectors a, b, c. Let |a| = 2, |b| = 3 and a = b × c. If α ∈ [0, π/3] is the angle bet

Q: Wait, why is $\vec{a} \cdot \vec{c} = 0$? I thought the dot product was only zero for orthogonal vectors.

You are absolutely correct! The dot product is zero if and only if the vectors are orthogonal. Since $\vec{a} = \vec{b} \times \vec{c}$, by the definition of the cross product, $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$. Therefore, $\vec{a} \perp \vec{c}$, which forces $\vec{a} \cdot \vec{c} = 0$. It is a beautiful geometric consequence.

Q: Why do we maximize $\sin^2 \alpha$ to minimize the expression? Shouldn't we minimize the denominator?

That is a great catch! Look at the expression $f(\alpha) = \frac{12}{\sin^2 \alpha} + 108$. Since $\sin^2 \alpha$ is in the denominator, as $\sin^2 \alpha$ increases, the fraction $\frac{12}{\sin^2 \alpha}$ decreases. To minimize the whole function, we need the largest possible value for the denominator.

Q: Why is the maximum of $\sin^2 \alpha$ at $\alpha = \frac{\pi}{3}$?

We are given the constraint $\alpha \in [0, \frac{\pi}{3}]$. In the first quadrant, the sine function is strictly increasing. Therefore, the function $\sin \alpha$ reaches its maximum value at the upper bound of the interval, which is $\frac{\pi}{3}$. Consequently, $\sin^2 \alpha$ also reaches its maximum at $\alpha = \frac{\pi}{3}$.

Question

Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}| = 2, |\vec{b}| = 3$ and $\vec{a} = \vec{b} 	imes \vec{c}$. If $\alpha \in [0, \frac{\pi}{3}]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c} - \vec{a}|^2$ is equal to:

Options:

(A) 110

(B) 124

(C) 121

(D) 105

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the world of JEE Advanced mathematics. Today, we are not just solving a problem; we are uncovering the hidden geometry behind vector operations.

We are given three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, with magnitudes $|\vec{a}| = 2$ and $|\vec{b}| = 3$, linked by the elegant relation $\vec{a} = \vec{b} 	imes \vec{c}$.

Our mission is to find the minimum value of $27|\vec{c} - \vec{a}|^2$ given that the angle $\alpha$ between $\vec{b}$ and $\vec{c}$ lies in the interval $[0, \frac{\pi}{3}]$.

### The Orthogonality Insight

The equation $\vec{a} = \vec{b} 	imes \vec{c}$ is more than just a formula; it is a geometric statement. By the very definition of the cross product, the resulting vector $\vec{a}$ must be perpendicular to the plane containing $\vec{b}$ and $\vec{c}$.

This implies that $\vec{a}$ is orthogonal to every vector in that plane, including $\vec{c}$ itself. In the language of dot products, this means:

$$\vec{a} \cdot \vec{c} = 0$$

This is our master key. Whenever you see a cross product, always look for this hidden orthogonality.

### The Algebraic Expansion

We turn our attention to the target expression: $27|\vec{c} - \vec{a}|^2$. We invoke the standard vector identity for the magnitude of a difference: $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$.

Applying this to our expression, we get:

$$27(|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{c})$$

Because we established that $\vec{a} \cdot \vec{c} = 0$, the term $-2\vec{a} \cdot \vec{c}$ vanishes into thin air. We are left with $27(|\vec{c}|^2 + |\vec{a}|^2)$. Since we know $|\vec{a}| = 2$, this simplifies to $27(|\vec{c}|^2 + 4)$.

### The Geometric Bridge

We are almost there, but we are missing $|\vec{c}|^2$. We return to our initial relation $\vec{a} = \vec{b} 	imes \vec{c}$ and take the magnitude of both sides:

$$|\vec{a}| = |\vec{b}| |\vec{c}| \sin \alpha$$

Substituting the known values, we have $2 = 3 |\vec{c}| \sin \alpha$. Solving for $|\vec{c}|$, we find:

$$|\vec{c}| = \frac{2}{3 \sin \alpha}$$

Squaring this gives us:

$$|\vec{c}|^2 = \frac{4}{9 \sin^2 \alpha}$$

### The Final Optimization

Now, we substitute this back into our simplified expression:

$$27 \left( \frac{4}{9 \sin^2 \alpha} + 4 ight)$$

Distributing the $27$, we get:

$$f(\alpha) = \frac{12}{\sin^2 \alpha} + 108$$

To minimize this function, we must maximize the denominator $\sin^2 \alpha$. Given $\alpha \in [0, \frac{\pi}{3}]$, the sine function is increasing, so the maximum occurs at $\alpha = \frac{\pi}{3}$.

Thus, $\sin^2(\frac{\pi}{3}) = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$. Substituting this back, we calculate:

$$\frac{12}{3/4} + 108 = 16 + 108 = 124$$

The **minimum value** of the expression is **124**. The elegance of the result is a testament to the harmony of vector algebra.

Animated Solution for Mathematics - Vector Algebra: Consider three vectors $a, b, c$ . Let $∣ a ∣ = 2, ∣ b ∣ = 3$ and $a = b \times c$ . If $α \in [0, \frac{π}{3}]$ is the angle between the vectors $b$ and $c$ , then the minimum value of $27∣ c - a ∣^{2}$ is equal to:

Select Answer:

Visualized Solution

Understanding $a = b \times c$

Orthogonality: $a \cdot c = 0$

Expanding $27∣ c - a ∣^{2}$

Simplifying the Expression

Using $∣ a ∣ = ∣ b \times c ∣$

Substituting Known Magnitudes

Isolating $∣ c ∣^{2}$

Formulating the Function $f (α)$

Minimizing the Function

Maximizing $sin^{2} α$

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Orthogonality Insight

The Algebraic Expansion

The Geometric Bridge

The Final Optimization

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Select Answer:

Visualized Solution

Understanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=b×c

Orthogonality: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅c=0

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 27∣c−a∣2