Let x, y and z be three vectors each of magnitude √2 and the angle between each pair of them is π/3.

Q: Why do we introduce a scalar $\lambda$ instead of just saying $\vec{a} = \vec{x} \times (\vec{y} \times \vec{z})$?

The cross product $\vec{x} \times (\vec{y} \times \vec{z})$ gives us the correct direction of $\vec{a}$, but not necessarily its magnitude. The problem states $\vec{a}$ is a non-zero vector, but it doesn't specify its length. Therefore, $\vec{a}$ could be any vector parallel to that cross product, which we represent by multiplying by a scalar $\lambda$.

Q: Why did we choose to dot with $\vec{y}$ to find $\lambda$?

It is a strategic choice! After expanding the triple product, we have $\vec{a} = \lambda(\vec{y} - \vec{z})$. If we dot with $\vec{y}$, we get $\vec{y} \cdot \vec{y}$ and $\vec{z} \cdot \vec{y}$, both of which we know (2 and 1, respectively). This makes the equation $\vec{a} \cdot \vec{y} = \lambda(2 - 1)$, which immediately gives $\lambda = \vec{a} \cdot \vec{y}$. It is the most efficient path to the solution.

Q: Is there a geometric way to see why $\vec{a} \cdot \vec{b} = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$?

While the algebraic derivation is the most robust method, geometrically, this result arises because the vectors $\vec{x}, \vec{y}, \vec{z}$ are symmetric in their arrangement (all at $\frac{\pi}{3}$ to each other). The negative sign emerges from the expansion of the dot product of the two resulting vectors, which effectively captures the geometric orientation of the vectors relative to each other.

Question

Let $\vec{x}, \vec{y}$ and $\vec{z}$ be three vectors each of magnitude $\sqrt{2}$ and the angle between each pair of them is $\frac{\pi}{3}$. If $\vec{a}$ is a non-zero vector perpendicular to $\vec{x}$ and $\vec{y} 	imes \vec{z}$ and $\vec{b}$ is a non-zero vector perpendicular to $\vec{y}$ and $\vec{z} 	imes \vec{x}$, then

Options:

(A) $\vec{b} = (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x})$

(B) $\vec{a} = (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z})$

(C) $\vec{a} \cdot \vec{b} = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$

(D) $\vec{a} = -(\vec{a} \cdot \vec{y})(\vec{z} - \vec{y})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE Advanced landscape. Today, we are not just solving a problem; we are dissecting the elegant geometry of vectors.

Imagine three vectors, $\vec{x}, \vec{y},$ and $\vec{z}$, each with a magnitude of $\sqrt{2}$, dancing in space, separated by an angle of $\frac{\pi}{3}$. This is our stage.

The dot product is the heartbeat of this problem. Since $|\vec{x}| = |\vec{y}| = |\vec{z}| = \sqrt{2}$ and the angle between any pair is $\frac{\pi}{3}$, we calculate:

$$\vec{x} \cdot \vec{y} = |\vec{x}||\vec{y}| \cos\left(\frac{\pi}{3}ight) = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1$$

This simple value of $1$ is the key that will unlock the entire puzzle.

### The Power of the Triple Product

Now, consider vector $\vec{a}$. We are told it is perpendicular to $\vec{x}$ and $\vec{y} 	imes \vec{z}$. Geometrically, if a vector is perpendicular to two others, it must be parallel to their cross product.

Thus, $\vec{a} \parallel \vec{x} 	imes (\vec{y} 	imes \vec{z})$. We write this as:

$$\vec{a} = \lambda [\vec{x} 	imes (\vec{y} 	imes \vec{z})]$$

This is where we summon the **Vector Triple Product identity**: $\vec{A} 	imes (\vec{B} 	imes \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$. Applying this to our expression, we get:

$$\vec{a} = \lambda [(\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}]$$

Substituting our known dot products, which are all $1$, the expression simplifies beautifully to $\vec{a} = \lambda (\vec{y} - \vec{z})$.

### The Scalar Hunt

We have a simplified expression, but we are haunted by the unknown scalar $\lambda$. How do we pin it down? We use the dot product as a probe.

By taking the dot product of $\vec{a} = \lambda (\vec{y} - \vec{z})$ with $\vec{y}$, we get:

$$\vec{a} \cdot \vec{y} = \lambda (\vec{y} \cdot \vec{y} - \vec{z} \cdot \vec{y})$$

Since $\vec{y} \cdot \vec{y} = |\vec{y}|^2 = 2$ and $\vec{z} \cdot \vec{y} = 1$, we find $\vec{a} \cdot \vec{y} = \lambda (2 - 1)$, which means $\lambda = \vec{a} \cdot \vec{y}$. Substituting this back, we arrive at:

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

This is a profound result, and it confirms **Option (B)**.

### The Final Synthesis

We apply the exact same logic to vector $\vec{b}$. Since $\vec{b}$ is perpendicular to $\vec{y}$ and $\vec{z} 	imes \vec{x}$, it is parallel to $\vec{y} 	imes (\vec{z} 	imes \vec{x})$.

Expanding this yields $\vec{b} = \mu (\vec{z} - \vec{x})$, and by dotting with $\vec{z}$, we find $\mu = \vec{b} \cdot \vec{z}$. Thus:

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

This confirms **Option (A)**. Finally, to check **Option (C)**, we compute the dot product $\vec{a} \cdot \vec{b}$:

$$\vec{a} \cdot \vec{b} = (\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z}) [(\vec{y} - \vec{z}) \cdot (\vec{z} - \vec{x})]$$

Expanding the vector part, we get:

$$\vec{y} \cdot \vec{z} - \vec{y} \cdot \vec{x} - \vec{z} \cdot \vec{z} + \vec{z} \cdot \vec{x} = 1 - 1 - 2 + 1 = -1$$

Substituting this back, we get $\vec{a} \cdot \vec{b} = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$. We have successfully navigated the vector maze, proving **Options (A), (B), and (C) are all correct**.

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Visualized Solution

$I ni t ia l S e t u p an d D o tP r o d u c t s$

$D i r ec t i o n o f V ec t or a$

$A ppl y in g V ec t or T r i pl e P r o d u c t$

$S im pl i f y in g V ec t or a$

$F in d in g t h e S c a l a r λ$

$E v a l u a t in g λ$

$F ina l E x p r ess i o n f or a$

$E x p r ess i o n f or V ec t or b$

$F in d in g t h e S c a l a r μ$

$C a l c u l a t in g a \cdot b$

$E x p an d in g t h eD o tP r o d u c t$

$F ina l C o n c l u s i o n$

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Power of the Triple Product

The Scalar Hunt

The Final Synthesis

Similar Questions

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Let $a, b$ and $c$ be three vectors such that $∣ a ∣ = 3, ∣ b ∣ = 5, b . c = 10$ and the angle between $b$ and $c$ is $\frac{π}{3}$ . If $a$ is perpendicular to vector $b \times c$ , then $∣ a \times (b \times c) ∣$ is equal to ________

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Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . Let $c$ be a vector such that $∣ c - a ∣ = 3$ , $∣ (a \times b) \times c ∣ = 3$ and the angle between $c$ and $a \times b$ be $3 0^{\circ}$ . Then $a \cdot c$ is equal to:

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } InitialSetupandDotProducts

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DirectionofVectora

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingVectora

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingtheScalarλ

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalExpressionfora

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ExpressionforVectorb

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingtheScalarμ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculatinga⋅b

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ExpandingtheDotProduct

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalConclusion