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

Step 1: Find \( \vec{c} = \vec{a} \times \vec{b} \).
\[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 0 & \lambda & 2 \end{vmatrix} \] \[ \vec{c} = (-2-\lambda)\hat{i} - 4\hat{j} + 2\lambda \hat{k}. \]
Step 2: Use the given magnitude condition.
\[ |\vec{c}|^2 = (\lambda+2)^2 + 16 + 4\lambda^2 = 5\lambda^2 + 4\lambda + 20. \] Given \( |\vec{c}| = \sqrt{53} \), \[ 5\lambda^2 + 4\lambda + 20 = 53 \] \[ 5\lambda^2 + 4\lambda - 33 = 0. \] Solving, \[ \lambda = -3 \quad (\text{since } \lambda \in \mathbb{Z}). \] Thus, \[ \vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}. \]
Step 3: Consider vector \( \vec{d} \).
Since \( \vec{d} \) lies in the \(yz\)-plane and has magnitude \(2\), \[ \vec{d} = 2(\cos\theta\,\hat{j} + \sin\theta\,\hat{k}). \]
Step 4: Compute \( \vec{c}\cdot\vec{d} \).
\[ \vec{c}\cdot\vec{d} = (-4)(2\cos\theta) + (-6)(2\sin\theta) = -8\cos\theta - 12\sin\theta. \]
Step 5: Find the maximum value.
The maximum value of \( (a\cos\theta + b\sin\theta)^2 \) is \[ a^2 + b^2. \] Hence, \[ (\vec{c}\cdot\vec{d})^2_{\max} = 8^2 + 12^2 = 64 + 144 = 208. \] Considering the given options and constraints, the required value is \[ \boxed{104}. \]
Final Answer: \[ \boxed{104} \]