Let a = 2i - j + k and b = λj + 2k, λ ∈ Z be two vectors. Let c = a × b and d be a vector of magnitu

Q: Why do we reject $\lambda = 2.2$?

The problem explicitly states $\lambda \in \mathbb{Z}$. In JEE Advanced, constraints are your best friends—they prune the solution space. When you see an integer constraint, treat it as a 'sanity check' that helps you verify if your quadratic roots are physically valid within the context of the problem.

Q: Can I use Lagrange Multipliers for the maximization?

You absolutely can! Lagrange Multipliers are a powerful, general-purpose tool. However, for a simple constraint like $y^2 + z^2 = 4$, the Cauchy-Schwarz inequality is the 'sniper' approach. It is faster, more elegant, and significantly less prone to algebraic errors under exam pressure.

Q: Why is the $x$-component of $\vec{d}$ zero?

Because $\vec{d}$ is defined to lie in the $yz$-plane. By definition, any vector residing in the $yz$-plane has no projection along the $x$-axis. Visualizing this geometric constraint is crucial; it simplifies your dot product calculation from a 3D problem to a 2D one immediately.

Question

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \lambda\hat{j} + 2\hat{k}, \lambda \in \mathbb{Z}$ be two vectors. Let $\vec{c} = \vec{a} 	imes \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $yz$-plane. If $|\vec{c}| = \sqrt{53}$, then the maximum possible value of $(\vec{c} \cdot \vec{d})^2$ is equal to :

Options:

(A) 26

(B) 52

(C) 208

(D) 104

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors, $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \lambda\hat{j} + 2\hat{k}$, where $\lambda$ is an integer. Our objective is to determine the maximum possible value of $(\vec{c} \cdot \vec{d})^2$, where $\vec{c} = \vec{a} 	imes \vec{b}$ and $\vec{d}$ is a vector of magnitude 2 lying in the $yz$-plane.

### The Determinant Dance

The cross product $\vec{c} = \vec{a} 	imes \vec{b}$ is calculated using the determinant:

$$
\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 0 & \lambda & 2 \end{vmatrix}
$$

Expanding this determinant, we find the components of $\vec{c}$:

$$
\vec{c} = \hat{i}((-1)(2) - (1)(\lambda)) - \hat{j}((2)(2) - (1)(0)) + \hat{k}((2)(\lambda) - (-1)(0))
$$

$$
\vec{c} = (-2 - \lambda)\hat{i} - 4\hat{j} + 2\lambda\hat{k}
$$

### The Integer Constraint

We are given that $|\vec{c}| = \sqrt{53}$, which implies $|\vec{c}|^2 = 53$. Substituting the components of $\vec{c}$ into the magnitude formula:

$$
(-2 - \lambda)^2 + (-4)^2 + (2\lambda)^2 = 53
$$

Expanding the terms, we obtain:

$$
(4 + \lambda^2 + 4\lambda) + 16 + 4\lambda^2 = 53
$$

$$
5\lambda^2 + 4\lambda - 33 = 0
$$

Solving this quadratic equation yields $\lambda = -3$ and $\lambda = 2.2$. Since the problem specifies that $\lambda \in \mathbb{Z}$, we must choose $\lambda = -3$. Substituting this value back into our expression for $\vec{c}$, we get:

$$
\vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}
$$

### The $yz$-Plane and the Dot Product

Since $\vec{d}$ lies in the $yz$-plane, its $x$-component is zero, so we define $\vec{d} = y\hat{j} + z\hat{k}$. Given $|\vec{d}| = 2$, we have the constraint $y^2 + z^2 = 4$.

The dot product $\vec{c} \cdot \vec{d}$ is:

$$
\vec{c} \cdot \vec{d} = (1)(0) + (-4)(y) + (-6)(z) = -4y - 6z
$$

We aim to maximize the square of this value, which is $(-4y - 6z)^2 = (4y + 6z)^2$.

### The Cauchy-Schwarz Masterclass

To maximize $(4y + 6z)^2$ subject to $y^2 + z^2 = 4$, we apply the **Cauchy-Schwarz Inequality**:

$$
(ay + bz)^2 \leq (a^2 + b^2)(y^2 + z^2)
$$

Setting $a = 4$ and $b = 6$, we substitute the known values:

$$
(4y + 6z)^2 \leq (4^2 + 6^2)(y^2 + z^2)
$$

$$
(4y + 6z)^2 \leq (16 + 36)(4)
$$

$$
(4y + 6z)^2 \leq (52)(4) = 208
$$

The maximum possible value of $(\vec{c} \cdot \vec{d})^2$ is **208**.

Select Answer:

Visualized Solution

Visualizing Vectors $a$ and $b$

Setting up the Cross Product $c = a \times b$

Expanding the Determinant

Using the Magnitude Condition $∣ c ∣ = 53$

Simplifying the Quadratic Equation

Solving for $λ$

Selecting the Valid $λ$

Defining Vector $d$ in the $y z$ -plane

Calculating the Dot Product $c \cdot d$

Applying Cauchy-Schwarz Inequality

Final Calculation of Maximum Value

Summary and Key Takeaways

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Determinant Dance

The Integer Constraint

The $y z$ -Plane and the Dot Product

The Cauchy-Schwarz Masterclass

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Select Answer:

Visualized Solution

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

Setting up the Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=a×b

Expanding the Determinant

Using the Magnitude Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣c∣=53​

Simplifying the Quadratic Equation

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

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

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d in the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yz-plane

Calculating the Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅d