Let a, b and c be three unit vectors such that a × (b × c) = √3/2 (b + c). If b is not parallel to c

Q: Why can we equate the coefficients of $\vec{b}$ and $\vec{c}$?

Because $\vec{b}$ and $\vec{c}$ are not parallel, they form a basis for the plane they span. Any vector in that plane can be uniquely represented as a linear combination of $\vec{b}$ and $\vec{c}$. Thus, the coefficients must match.

Q: Why is the dot product $\vec{a} \cdot \vec{b}$ negative?

The dot product is negative when the angle between the vectors is obtuse (greater than $90^\circ$). Since $\cos \theta = -\frac{\sqrt{3}}{2}$, $\theta$ must be in the second quadrant, specifically $150^\circ$ or $\frac{5\pi}{6}$ radians.

Q: Can we use the coefficient of $\vec{b}$ instead?

Yes, you can! Equating the coefficients of $\vec{b}$ gives $\vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2}$. This tells us the angle between $\vec{a}$ and $\vec{c}$ is $\frac{\pi}{6}$. However, the question specifically asks for the angle between $\vec{a}$ and $\vec{b}$, so focusing on the $\vec{c}$ coefficient is the direct path.

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c})$. If $\vec{b}$ is not parallel to $\vec{c}$, then the angle between $\vec{a}$ and $\vec{b}$ is:

Options:

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

(B) $\frac{5\pi}{6}$

(C) $\frac{3\pi}{4}$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system, holding three unit vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. They are bound by the relationship:

$$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c})$$

To find the angle $	heta$ between $\vec{a}$ and $\vec{b}$, we must first dismantle the vector triple product on the left side using the **'BAC-CAB' rule**.

The identity states that:

$$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$

By substituting this identity into our original equation, we transform the expression into a linear combination of $\vec{b}$ and $\vec{c}$:

$$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\sqrt{3}}{2}\vec{b} + \frac{\sqrt{3}}{2}\vec{c}$$

### The Power of Linear Independence

We now have a linear combination of $\vec{b}$ and $\vec{c}$ on both sides of the equation. Because the problem implies that $\vec{b}$ and $\vec{c}$ are not parallel, they are **linearly independent**.

In the language of linear algebra, they form a basis for the plane they span. Consequently, we can equate the coefficients of $\vec{b}$ and $\vec{c}$ on both sides.

Comparing the coefficients of $\vec{c}$, we obtain:

$$-(\vec{a} \cdot \vec{b}) = \frac{\sqrt{3}}{2}$$

This immediately yields the dot product:

$$\vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2}$$

### The Geometric Bridge

We have successfully extracted the algebraic value of the dot product. Recall the fundamental definition of the dot product:

$$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$$

Since $\vec{a}$ and $\vec{b}$ are unit vectors, their magnitudes are both $1$. Thus, the equation simplifies to:

$$\cos 	heta = -\frac{\sqrt{3}}{2}$$

We are looking for an angle $	heta$ in the range $[0, \pi]$. Since the cosine is negative, the angle must be obtuse and lie in the second quadrant.

Knowing that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, we calculate the angle as:

$$	heta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$

Through the power of the vector triple product and the principle of linear independence, we have uncovered the hidden angle. The final result is:

$$	heta = \frac{5\pi}{6}$$

Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be three unit vectors such that $a \times (b \times c) = \frac{3}{2} (b + c)$ . If $b$ is not parallel to $c$ , then the angle between $a$ and $b$ is:

Select Answer:

Visualized Solution

Visualizing the Vectors

The Given Equation

Vector Triple Product (VTP)

Substituting the VTP

Expanding the Right Side

Linear Independence

Comparing Coefficients of $c$

Isolating the Dot Product

Dot Product Formula

Substituting Magnitudes

Solving for $cos θ$

Finding the Final Angle

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Power of Linear Independence

The Geometric Bridge

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