Let a, b and c be three unit vectors, out of which vectors b and c are non-parallel. If α and β are

Q: Why can we just equate the coefficients of $\vec{b}$ and $\vec{c}$? Is that always allowed?

Great question! We can only equate coefficients if the vectors are linearly independent. Since the problem explicitly states $\vec{b}$ and $\vec{c}$ are non-parallel, they are linearly independent. This means any vector in their plane has a unique representation as a linear combination of $\vec{b}$ and $\vec{c}$. If they were parallel, the representation would not be unique, and we couldn't simply compare coefficients.

Q: I keep forgetting the BAC-CAB rule. Is there a trick?

Think of it as 'BAC minus CAB'. The vector in the middle of the cross product (the one being crossed with the result of the other cross) is the one that gets dotted with the outer vector first. It is a classic identity, and once you derive it or use it enough, it becomes second nature. Just remember: the vector that is 'crossed' twice (the one in the middle) is the one that gets the negative sign in the expansion.

Q: What if the dot product resulted in a value greater than 1? Would that be a problem?

Absolutely! Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, their dot product must lie in the interval $[-1, 1]$. If you ever get a value outside this range, it is a red flag that you have made a calculation error. Always check your bounds!

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors, out of which vectors $\vec{b}$ and $\vec{c}$ are non-parallel. If $\alpha$ and $\beta$ are the angles which vector $\vec{a}$ makes with vectors $\vec{b}$ and $\vec{c}$ respectively and $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{1}{2}\vec{b}$, then $|\alpha - \beta|$ is equal to :

Options:

(A) $60^\circ$

(B) $30^\circ$

(C) $90^\circ$

(D) $45^\circ$

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into the Triple Product

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric puzzle.

When you look at a vector equation like $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{1}{2}\vec{b}$, do not see it as a dry collection of symbols. See it as a conversation between three arrows in space.

We are given three unit vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$. The fact that they are unit vectors is a gift—it means their magnitudes are unity, simplifying our dot products into pure cosines. But the real magic lies in the vector triple product.

### Phase 1: The BAC-CAB Identity

The expression $\vec{a} 	imes (\vec{b} 	imes \vec{c})$ is a classic in vector algebra. It is the vector triple product.

If you have ever felt intimidated by this, take a deep breath. There is a mnemonic that has saved countless students: the 'BAC-CAB' rule. The expansion is:

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

Geometrically, the cross product $\vec{b} 	imes \vec{c}$ creates a vector perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. When we cross $\vec{a}$ with that result, we are essentially rotating back into the plane of $\vec{b}$ and $\vec{c}$. That is why the result is a linear combination of $\vec{b}$ and $\vec{c}$.

### Phase 2: The Power of Linear Independence

Now, look at the equation again: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b}$. We are told that $\vec{b}$ and $\vec{c}$ are non-parallel.

This is the key that unlocks the door. In the world of linear algebra, if two vectors are non-parallel, they are linearly independent. This means they form a basis for the plane they span.

Because the representation is unique, we can compare the coefficients on both sides of the equation. On the right side, we have $\frac{1}{2}\vec{b} + 0\vec{c}$. On the left side, we have $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.

By matching the coefficients, we get two simple, beautiful equations:

$$ \vec{a} \cdot \vec{c} = \frac{1}{2} \quad 	ext{and} \quad \vec{a} \cdot \vec{b} = 0 $$

### Phase 3: Connecting to Angles

We are almost there. We know that for any two vectors $\vec{u}$ and $\vec{v}$, the dot product is defined as $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos 	heta$.

Since our vectors are unit vectors, $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Therefore, the dot product is simply the cosine of the angle between them.

For $\vec{a} \cdot \vec{c} = \frac{1}{2}$, we have $\cos \beta = \frac{1}{2}$. This tells us that $\beta = 60^\circ$.

For $\vec{a} \cdot \vec{b} = 0$, we have $\cos \alpha = 0$, which means $\alpha = 90^\circ$. The vectors $\vec{a}$ and $\vec{b}$ are perpendicular!

### Conclusion: The Final Step

The problem asks for $|\alpha - \beta|$. We have $\alpha = 90^\circ$ and $\beta = 60^\circ$.

The difference is $|90^\circ - 60^\circ| = \mathbf{30^\circ}$.

You see? By trusting the identity and respecting the geometric constraints, we navigated through the algebra to find a clean, precise answer. Keep this mindset—always look for the geometric meaning behind the algebraic manipulation.

Select Answer:

Visualized Solution

Unit Vectors Setup

Angles $α$ and $β$

Vector Triple Product

Expanding the Equation

Comparing Coefficients

Equating Coefficients of $b$

Equating Coefficients of $c$

Applying Dot Product Formula

Finding Angle $β$

Finding Angle $α$

Calculating $∣ α - β ∣$

The Sigma Insight: Vector Triple Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Conclusion

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