Let a = i + j + k, b = 3i + 2j - k, c = λj + μk and d be a unit vector such that a × d = b × d and c

Q: Wait, if $(\vec{a} - \vec{b}) \times \hat{d} = 0$, does that mean $\vec{a} = \vec{b}$?

Not at all! A cross product being zero means the vectors are parallel, not necessarily equal. It means $\hat{d}$ is a scalar multiple of $(\vec{a} - \vec{b})$. Since $\hat{d}$ is a unit vector, it is simply the normalized version of $(\vec{a} - \vec{b})$.

Q: Why do we have a $\pm$ sign for $\hat{d}$? Does it change the final answer?

Great observation. The $\pm$ sign appears because $\hat{d}$ could point in either direction along the line defined by $(\vec{a} - \vec{b})$. However, when you solve for $\lambda$ using $\vec{c} \cdot \hat{d} = 1$, the sign of $\lambda$ adjusts to compensate. When you square the final expression, the sign effectively vanishes, ensuring a unique, consistent answer.

Q: Why is $\vec{c} \cdot \hat{d} = 1$ so important?

It is the key constraint! Without this, we would have $\lambda$ and $\mu$ as unknowns without a way to fix their magnitude. This dot product locks the geometry in place, allowing us to solve for the specific values of $\lambda$ and $\mu$.

Question

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 3\hat{i} + 2\hat{j} - \hat{k}, \vec{c} = \lambda\hat{j} + \mu\hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} 	imes \hat{d} = \vec{b} 	imes \hat{d}$ and $\vec{c} \cdot \hat{d} = 1$. If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3\lambda\hat{d} + \mu\vec{c}|^2$ is equal to _______ .

Sigma Percentile · Accepted Answer

### The Geometry of Space: Unlocking the Vector Puzzle

Welcome, fellow traveler in the realm of JEE Advanced mathematics. Today, we are not just solving a problem; we are decoding the language of three-dimensional space.

We are given vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, and a mysterious unit vector $\hat{d}$. At first glance, this looks like a chaotic mess of symbols, but let us peel back the layers together.

### Phase 1: The Cross Product Insight

We start with the equation $\vec{a} 	imes \hat{d} = \vec{b} 	imes \hat{d}$. If we bring everything to one side, we get:

$$(\vec{a} - \vec{b}) 	imes \hat{d} = \vec{0}$$

In the world of vectors, a cross product of zero is a massive signal. It tells us that the vector $(\vec{a} - \vec{b})$ and the unit vector $\hat{d}$ are perfectly parallel.

By calculating $\vec{a} - \vec{b}$, we find:

$$\vec{a} - \vec{b} = (1-3)\hat{i} + (1-2)\hat{j} + (1-(-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$$

Since $\hat{d}$ is a unit vector, we divide by its magnitude, which is $\sqrt{(-2)^2 + (-1)^2 + 2^2} = 3$. Thus, we have:

$$\hat{d} = \pm \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k})$$

### Phase 2: The Orthogonality Constraint

Now, let us look at $\vec{c} = \lambda\hat{j} + \mu\hat{k}$. We are told $\vec{c} \perp \vec{a}$, which means the dot product must vanish: $\vec{c} \cdot \vec{a} = 0$.

Substituting the components, we get:

$$(\lambda\hat{j} + \mu\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$$

This simplifies beautifully to $\lambda(1) + \mu(1) = 0$, or $\mu = -\lambda$. Our vector $\vec{c}$ is now constrained to $\lambda(\hat{j} - \hat{k})$.

### Phase 3: The Final Connection

We have one last piece of the puzzle: $\vec{c} \cdot \hat{d} = 1$. This is the bridge that fixes our variables.

Substituting our expressions for $\vec{c}$ and $\hat{d}$:

$$\lambda(\hat{j} - \hat{k}) \cdot \left[ \pm \frac{1}{3}(-2\hat{i} - \hat{j} + 2\hat{k}) ight] = 1$$

Calculating the dot product, the $\hat{j}$ components give $(-1)$ and the $\hat{k}$ components give $(-2)$. The sum is $-3$. So:

$$\pm \frac{\lambda}{3}(-3) = 1 \Rightarrow \mp \lambda = 1 \Rightarrow \lambda = \mp 1$$

### Phase 4: The Grand Finale

Finally, we need to evaluate $|3\lambda\hat{d} + \mu\vec{c}|^2$. Using the expansion $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2(\vec{u} \cdot \vec{v})$, we get:

$$9\lambda^2|\hat{d}|^2 + \mu^2|\vec{c}|^2 + 6\lambda\mu(\hat{d} \cdot \vec{c})$$

Since $\lambda^2 = 1$, $\mu^2 = 1$, $|\vec{c}|^2 = \lambda^2 + \mu^2 = 2$, and $\hat{d} \cdot \vec{c} = 1$, the expression becomes:

$$9(1)(1) + (1)(2) + 6(\lambda\mu)(1)$$

Since $\lambda\mu = -\lambda^2 = -1$, we have $9 + 2 - 6 = 5$. The elegance of the cancellation is truly satisfying.

The final answer is **5**.

Enter Numerical Value:

Visualized Solution

Analyzing $a \times \hat{d} = b \times \hat{d}$

Calculating $a - b$

Finding Unit Vector $\hat{d}$

Orthogonality Condition $c ⊥ a$

Using $c \cdot \hat{d} = 1$

Expanding $∣3 λ \hat{d} + μ c ∣^{2}$

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Space

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a×d^=b×d^

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

Finding Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d^

Orthogonality Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⊥a

Using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅d^=1

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣3λd^+μc∣2