If (a × b) × c = a × (b × c) where a, b and c are any three vectors such that a · b ≠ 0, b · c ≠ 0 t

Q: Wait, why can we just cancel the $(\vec{a} \cdot \vec{c})\vec{b}$ term? Is it always allowed?

Yes, absolutely! In vector algebra, we are dealing with equality of vectors. If you have an equation of the form $\vec{X} + \vec{Y} = \vec{X} + \vec{Z}$, you can subtract $\vec{X}$ from both sides to get $\vec{Y} = \vec{Z}$. Here, $(\vec{a} \cdot \vec{c})\vec{b}$ is a vector (a scalar times a vector), so it behaves exactly like any other vector in an equation. It is not division; it is vector subtraction.

Q: Why can't we just say $\vec{a} = \vec{c}$? Is $\lambda$ always 1?

That is a great question! The relation we derived is $\vec{a} = \lambda \vec{c}$, where $\lambda = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{c}}$. This $\lambda$ is a scalar, but it is not necessarily 1. It depends on the specific vectors $\vec{a}, \vec{b},$ and $\vec{c}$. The definition of parallel vectors is that one is a scalar multiple of the other, so $\vec{a} = \lambda \vec{c}$ is the general condition for parallelism, regardless of the value of $\lambda$.

Q: What if $\vec{b} \cdot \vec{c} = 0$? Does the problem break?

The problem explicitly states that $\vec{b} \cdot \vec{c} \neq 0$. This constraint is crucial because it ensures that we are not dividing by zero when we isolate $\vec{a}$. If $\vec{b} \cdot \vec{c}$ were zero, the equation would become $0 = -(\vec{a} \cdot \vec{b})\vec{c}$, which would imply either $\vec{a} \cdot \vec{b} = 0$ or $\vec{c} = \vec{0}$. The problem avoids this ambiguity by providing the non-zero constraint.

Question

If $(\vec{a} 	imes \vec{b}) 	imes \vec{c} = \vec{a} 	imes (\vec{b} 	imes \vec{c})$ where $\vec{a}, \vec{b}$ and $\vec{c}$ are any three vectors such that $\vec{a} \cdot \vec{b} 
eq 0, \vec{b} \cdot \vec{c} 
eq 0$ then $\vec{a}$ and $\vec{c}$ are

Options:

(A) inclined at an angle of $\pi/3$ between them

(B) inclined at an angle of $\pi/6$ between them

(C) perpendicular

(D) parallel

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space, surrounded by three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. These are the fundamental building blocks of spatial geometry.

We are investigating the condition of associativity for the vector triple product:
$$(\vec{a} 	imes \vec{b}) 	imes \vec{c} = \vec{a} 	imes (\vec{b} 	imes \vec{c})$$

At first glance, this looks like a simple algebraic identity, but it hides a profound geometric truth regarding the relationship between $\vec{a}$ and $\vec{c}$.

### The BAC-CAB Rule

To solve this, we utilize the vector triple product expansion, commonly known as the **BAC-CAB rule**. This rule allows us to convert a cross product of cross products into a linear combination of the vectors involved.

For the left-hand side (LHS), the rule states:
$$(\vec{a} 	imes \vec{b}) 	imes \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$$

For the right-hand side (RHS), applying the same logic with careful attention to the order of vectors, we obtain:
$$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$

### The Moment of Cancellation

Now, we equate the two sides of the expression:
$$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$

Notice that the term $(\vec{a} \cdot \vec{c})\vec{b}$ appears on both sides. We can subtract this term from both sides to simplify the equation:
$$-(\vec{b} \cdot \vec{c})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c}$$

Multiplying the entire equation by $-1$, we arrive at the simplified relationship:
$$(\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c}$$

### The Geometric Revelation

Given the condition that $\vec{b} \cdot \vec{c} 
eq 0$, we can divide both sides by the scalar $(\vec{b} \cdot \vec{c})$:
$$\vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{c}} ight) \vec{c}$$

Let the scalar term in the parentheses be represented by $\lambda$. Thus, we have:
$$\vec{a} = \lambda \vec{c}$$

In vector algebra, if one vector is a scalar multiple of another, they must be **parallel**. We have successfully stripped away the complexity to reveal that **$\vec{a}$ and $\vec{c}$ are parallel**.

Animated Solution for Mathematics - Vector Algebra: If $(a \times b) \times c = a \times (b \times c)$ where $a, b$ and $c$ are any three vectors such that $a \cdot b \neq = 0, b \cdot c \neq = 0$ then $a$ and $c$ are

Select Answer:

Visualized Solution

The Given Condition

Expanding the LHS

Expanding the RHS

Equating LHS and RHS

Canceling Common Terms

Handling the Negative Signs

Isolating Vector $a$

The Scalar Multiplier

Conclusion: Parallel Vectors

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The BAC-CAB Rule

The Moment of Cancellation

The Geometric Revelation

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (a×b)×c=a×(b×c) where a,b and c are any three vectors such that a⋅b=0,b⋅c=0 then a and c are