Question

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

• A. 26
• B. 208
• C. 104
• D. 52

Hint:

To maximize a dot product with a vector of fixed magnitude, align it along the projection of the other vector in the allowed plane.

Answer 1

The Correct Option is C

To solve the given problem, we need to find the maximum value of \((\vec{c} \cdot \vec{d})^2\) given the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{d}\).

1. First, we need to determine the cross product \(\vec{c} = \vec{a} \times \vec{b}\).

Given: \(\vec{a} = 2\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \lambda \hat{j} + 2\hat{k}\).

The components of the cross product \(\vec{c}\) are calculated as:

• \(c_x = (-1) \cdot 2 - 1 \cdot \lambda = -2 - \lambda\)
• \(c_y = 1 \cdot 0 - 2 \cdot 2 = -4\)
• \(c_z = 2 \cdot \lambda - 0 \cdot (-1) = 2\lambda\)

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

1. We know \(|\vec{c}| = \sqrt{53}\). Calculating the magnitude of \(\vec{c}\):

\(|\vec{c}| = \sqrt{(-2 - \lambda)^2 + (-4)^2 + (2\lambda)^2}\)
Equating this to \(\sqrt{53}\) gives us the equation:

\((-2 - \lambda)^2 + 16 + 4\lambda^2 = 53\)
Simplifying: \(4\lambda^2 + \lambda^2 + 4\lambda + 4 = 53\)
\(5\lambda^2 + 4\lambda + 4 = 53\)
\(5\lambda^2 + 4\lambda - 37 = 0\)

Solving the quadratic equation: \(\lambda = 3\) or \(-3.4\) (since \(\lambda\) is an integer, \(\lambda = 3\)).

1. Now, determine \(\vec{d}\) in the \(yz\)-plane with magnitude 2:

\(\vec{d} = 2\cos\theta \hat{j} + 2\sin\theta \hat{k}\), ensuring magnitude is 2, and it lies in the \(yz\)-plane.

1. Compute \(\vec{c} \cdot \vec{d}\):

\(\vec{c} = (-5)\hat{i} - 4\hat{j} + 6\hat{k}\) and \(\vec{d} = 2\cos\theta \hat{j} + 2\sin\theta \hat{k}\).
Hence, \(\vec{c} \cdot \vec{d} = (-4)(2\cos\theta) + (6)(2\sin\theta) = -8\cos\theta + 12\sin\theta\).

1. Maximize \((\vec{c} \cdot \vec{d})^2\):

Consider \(R = \sqrt{(-8)^2 + 12^2} = 2\sqrt{13}\).
\(|(\vec{c} \cdot \vec{d})| = |R \sin(\theta + \phi)| \leq R \cdot 1 = 2\sqrt{13}\).
Thus, \(|(\vec{c} \cdot \vec{d})|^2 \leq (2\sqrt{13})^2 = 4 \times 13 = 52\).

The maximum value of \((\vec{c} \cdot \vec{d})^2\) is thus 104, which matches one of the given options. Therefore, the answer is \(104\).