Let a, b and c be three non-zero vectors such that no two of them are collinear and (a × b) × c = 1/

Q: Wait, why can we just set the coefficients of $\vec{a}$ and $\vec{b}$ to zero? Is that always true?

That is a fantastic question! In vector algebra, if $\vec{a}$ and $\vec{b}$ are non-collinear, they are linearly independent. This means the only way to satisfy $x\vec{a} + y\vec{b} = \vec{0}$ is if $x=0$ and $y=0$. If they were collinear, one could be written as a multiple of the other, and the coefficients wouldn't necessarily have to be zero. Since the problem explicitly states they are non-collinear, we are safe to equate the coefficients to zero.

Q: Why did the $\vec{a}$ term on the right side of the original equation not disappear?

It doesn't disappear; it gets absorbed! When we expand the triple product, we get a term involving $\vec{a}$ and a term involving $\vec{b}$. By moving everything to one side, we group all terms attached to $\vec{a}$ together. The right-hand side of the original equation is just a scalar multiple of $\vec{a}$, so it simply adds to the coefficient of $\vec{a}$ we already had from the expansion.

Q: Can $\sin \theta$ be negative? The options have negative values.

In the context of the angle $\theta$ between two vectors, $\theta$ is defined in the range $[0, \pi]$. Within this range, the sine function is always non-negative. While the cosine can be negative (as we found, $\cos \theta = -1/3$), the sine must remain positive. If you ever see a negative sine for an angle between vectors, double-check your definition of $\theta$!

Q: Why did we ignore the coefficient of $\vec{b}$? Does it have to be zero too?

Yes, it absolutely does! In the equation $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c} + \frac{1}{3} |\vec{b}| |\vec{c}|)\vec{a} = \vec{0}$, the coefficient of $\vec{b}$ is $(\vec{a} \cdot \vec{c})$. For the equation to hold, this must be zero. This implies $\vec{a}$ is perpendicular to $\vec{c}$. While we didn't need this to find $\sin \theta$, it is a vital piece of the geometric puzzle!

Q: Is there a geometric way to see this without the triple product identity?

You could visualize it, but it would be incredibly difficult! The triple product identity is the 'algebraic shortcut' that encodes the geometry of the cross product. Trying to solve this purely by drawing vectors would require you to decompose $\vec{c}$ into components parallel and perpendicular to the plane of $\vec{a}$ and $\vec{b}$, which is exactly what the identity does for you automatically.

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}$. If $	heta$ is the angle between vectors $\vec{b}$ and $\vec{c}$, then a value of $\sin 	heta$ is :

Options:

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

(B) $\frac{-2\sqrt{3}}{3}$

(C) $\frac{2\sqrt{2}}{3}$

(D) $\frac{-\sqrt{2}}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space, holding three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. They are constrained by a rigid relationship:

$$(\vec{a} 	imes \vec{b}) 	imes \vec{c} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}$$

At first glance, this looks like a chaotic mess of cross products and magnitudes. However, behind every complex equation lies a hidden symmetry waiting to be uncovered.

### The Power of the Identity

When dealing with a vector triple product, your first instinct should be to reach for the **Vector Triple Product Identity**. It is the master key that unlocks the geometry of cross products. The identity states:

$$\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$$

Our expression is $(\vec{a} 	imes \vec{b}) 	imes \vec{c}$. By noting that $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = -\vec{c} 	imes (\vec{a} 	imes \vec{b})$, we apply the identity to obtain:

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

This transformation converts a cross product into a simple linear combination of $\vec{a}$ and $\vec{b}$.

### The Dance of Coefficients

Now, we equate our expanded form to the right-hand side provided in the problem:

$$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = \frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}$$

Rearranging the terms to one side reveals the underlying structure:

$$(\vec{a} \cdot \vec{c})\vec{b} - \left( \vec{b} \cdot \vec{c} + \frac{1}{3} |\vec{b}| |\vec{c}| ight) \vec{a} = \vec{0}$$

Because $\vec{a}$ and $\vec{b}$ are non-collinear, they are **linearly independent**. For the equation to equal the zero vector, the coefficient of each vector must independently vanish.

This implies that $(\vec{a} \cdot \vec{c}) = 0$, meaning $\vec{a}$ is perpendicular to $\vec{c}$. Furthermore, we obtain the critical relationship:

$$\vec{b} \cdot \vec{c} + \frac{1}{3} |\vec{b}| |\vec{c}| = 0$$

### The Final Trigonometric Reveal

We know that the dot product is defined as $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 	heta$. Substituting this into our equation yields:

$$|\vec{b}| |\vec{c}| \cos 	heta = -\frac{1}{3} |\vec{b}| |\vec{c}|$$

Assuming non-zero magnitudes, we divide both sides by $|\vec{b}| |\vec{c}|$ to find $\cos 	heta = -\frac{1}{3}$. To find $\sin 	heta$, we use the identity $\sin^2 	heta + \cos^2 	heta = 1$:

$$\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \left( -\frac{1}{3} ight)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}}$$

Simplifying this, we arrive at the final result:

$$\sin 	heta = \frac{2\sqrt{2}}{3}$$

### Reflection

We started with a daunting vector equation and, by applying the correct identity and respecting the geometric constraints of non-collinearity, we distilled it into a precise trigonometric value. This is the beauty of mathematics—it is not about memorizing formulas, but about recognizing the structure within the chaos. You have successfully navigated the vector space.

Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be three non-zero vectors such that no two of them are collinear and $(a \times b) \times c = \frac{1}{3} ∣ b ∣∣ c ∣ a$ . If $θ$ is the angle between vectors $b$ and $c$ , then a value of $sin θ$ is :

Select Answer:

Visualized Solution

Visualizing the Vectors

The Vector Triple Product

Applying the VTP Identity

Rearranging the Equation

The Logic of Non-Collinearity

Equating Coefficients to Zero

Expanding the Dot Product

Solving for Cosine Theta

Finding Sine Theta

Final Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Power of the Identity

The Dance of Coefficients

The Final Trigonometric Reveal

Reflection

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