If a, b, c are three non-zero vectors and n is a unit vector perpendicular to c such that a = αb - n

Q: Wait, why can't we just solve for $\vec{a}$ and $\vec{b}$ individually? The problem doesn't give us their components!

That is the beauty of vector algebra! In JEE Advanced, you are rarely expected to find the individual components of vectors unless explicitly asked. We are working with the properties of the vectors. By using the Vector Triple Product expansion, we manipulate the relationships between them, allowing the unknown components to cancel out elegantly.

Q: I always get confused with the 'BAC minus CAB' rule. How do I know which vector is which?

Think of it as a mnemonic. For $\vec{A} \times (\vec{B} \times \vec{C})$, the result is $(\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$. The vector outside the parenthesis (the 'A') dots with the vector furthest away first (the 'C'), then the vector closer (the 'B'). Just remember: 'Outside-Far-Near' minus 'Outside-Near-Far'.

Q: Why is the final answer just 12? What happened to the $\alpha$ and the negative sign?

Great observation! When we substituted $\vec{c} \cdot \vec{a} = 12\alpha$ into the expression, we got $12\vec{a} - 12\alpha\vec{b}$. Factoring out the 12 gave us $12(\vec{a} - \alpha\vec{b})$. Since $\vec{a} - \alpha\vec{b} = -\hat{n}$, the expression became $-12\hat{n}$. When we take the magnitude, the negative sign disappears, and since $|\hat{n}| = 1$, we are left with 12. The $\alpha$ terms effectively cancelled out because they were part of the linear relationship defining $\vec{a}$.

Question

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero vectors and $\hat{n}$ is a unit vector perpendicular to $\vec{c}$ such that $\vec{a} = \alpha \vec{b} - \hat{n}, (\alpha 
eq 0)$ and $\vec{b} \cdot \vec{c} = 12$, then $|\vec{c} 	imes (\vec{a} 	imes \vec{b})|$ is equal to:

Options:

(A) 15

(B) 9

(C) 12

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of vector algebra. When you first look at a problem involving $\vec{a}$, $\vec{b}$, $\vec{c}$, and a mysterious unit vector $\hat{n}$, it is natural to feel a slight tremor of hesitation.

In JEE Advanced, the most beautiful solutions rarely come from brute force; they come from understanding the structural relationships between the entities. Let us dissect this problem with the precision of a surgeon and the curiosity of an explorer.

### The Geometry of the Hidden

We begin with the given relation:
$$\vec{a} = \alpha \vec{b} - \hat{n}$$

This is not just an equation; it is a geometric instruction. It tells us that $\vec{a}$ is a linear combination of $\vec{b}$ and $\hat{n}$.

We are told $\hat{n}$ is a unit vector perpendicular to $\vec{c}$. In the language of vectors, perpendicularity is synonymous with a dot product of zero. So, immediately, we know that:
$$\hat{n} \cdot \vec{c} = 0$$

### The Vector Triple Product Strategy

Now, we turn our gaze to the objective: finding the magnitude of $\vec{c} 	imes (\vec{a} 	imes \vec{b})$. When you see a cross product nested inside another, your intuition should scream "Vector Triple Product!"

We invoke the expansion formula:
$$\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$$

Applying this to our specific vectors, we get:
$$\vec{c} 	imes (\vec{a} 	imes \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$$

We are given $\vec{b} \cdot \vec{c} = 12$. Substituting this, our expression becomes:
$$12\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$$

### The Algebraic Dance

We return to our original equation: $\vec{a} = \alpha \vec{b} - \hat{n}$. If we want to find the dot product of $\vec{a}$ with $\vec{c}$, we simply dot both sides of this equation with $\vec{c}$.

$$\vec{a} \cdot \vec{c} = (\alpha \vec{b} - \hat{n}) \cdot \vec{c}$$

Distributing the dot product, we get:
$$\vec{a} \cdot \vec{c} = \alpha(\vec{b} \cdot \vec{c}) - (\hat{n} \cdot \vec{c})$$

Since $\hat{n} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{c} = 12$, we find:
$$\vec{a} \cdot \vec{c} = \alpha(12) - 0 = 12\alpha$$

### The Elegant Conclusion

Now, we substitute $\vec{c} \cdot \vec{a} = 12\alpha$ back into our expanded triple product expression:
$$12\vec{a} - (12\alpha)\vec{b} = 12(\vec{a} - \alpha\vec{b})$$

From the original equation $\vec{a} = \alpha \vec{b} - \hat{n}$, we rearrange to find $\vec{a} - \alpha\vec{b} = -\hat{n}$. Our expression simplifies to:
$$12(-\hat{n}) = -12\hat{n}$$

The question asks for the magnitude:
$$|\vec{c} 	imes (\vec{a} 	imes \vec{b})| = |-12\hat{n}| = 12|\hat{n}|$$

Since the magnitude of a unit vector is 1, the final result is **12**. You have navigated the complexity and arrived at the truth.

Animated Solution for Mathematics - Vector Algebra: If $a, b, c$ are three non-zero vectors and $\overset{n}{^}$ is a unit vector perpendicular to $c$ such that $a = α b - \overset{n}{^}, (α \neq = 0)$ and $b \cdot c = 12$ , then $∣ c \times (a \times b) ∣$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Base Vectors

The Linear Combination

The Objective: Vector Triple Product

Applying the VTP Formula

Substituting Known Values

Finding the Missing Dot Product

Expanding the Dot Product

Evaluating the Dot Products

Substituting Back into VTP

Factoring and Simplifying

Final Magnitude Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Geometry of the Hidden

The Vector Triple Product Strategy

The Algebraic Dance

The Elegant Conclusion

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