Question

Let $\vec{a} = \hat{i} + 4\hat{j}$, $\vec{b} = 4\hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - 2\hat{k}$. If $\vec{d}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ such that $\vec{c} \cdot \vec{d} = 16$, then $|\vec{d}|$ is equal to

• A. $\sqrt{33}$
• B. $2\sqrt{33}$
• C. $3\sqrt{33}$
• D. $4\sqrt{33}$

Hint:

The cross product $\vec{a} \times \vec{b}$ gives a vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. Any vector perpendicular to this plane must be a scalar multiple of the cross product. This is a fundamental concept for solving such problems.

Answer 1

The Correct Option is D

Step 1: Conceptual Foundation:
A vector orthogonal to both $\vec{a}$ and $\vec{b}$ is collinear with their cross product, $\vec{a} \times \vec{b}$. Consequently, $\vec{d}$ can be represented as a scalar multiple of this cross product. The scalar is determined using the provided dot product condition, after which the magnitude of $\vec{d}$ is computed.
Step 2: Methodological Outline:
1. Compute the cross product $\vec{n} = \vec{a} \times \vec{b}$.2. Define $\vec{d} = \lambda \vec{n}$, where $\lambda$ is a scalar.3. Solve for $\lambda$ using the constraint $\vec{c} \cdot \vec{d} = 16$.4. Calculate the magnitude $|\vec{d}|$.
Step 3: Procedural Breakdown:
Given vectors: $\vec{a} = \hat{i} + 4\hat{j} + 0\hat{k}$ and $\vec{b} = 0\hat{i} + 4\hat{j} + \hat{k}$.
Calculate the cross product $\vec{a} \times \vec{b}$:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 4 & 0
0 & 4 & 1 \end{vmatrix} = \hat{i}(4 \cdot 1 - 0 \cdot 4) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 4 - 0 \cdot 0) \]\[ \vec{a} \times \vec{b} = 4\hat{i} - \hat{j} + 4\hat{k} \]Since $\vec{d}$ is perpendicular to $\vec{a}$ and $\vec{b}$, it is parallel to $\vec{a} \times \vec{b}$.
Let $\vec{d} = \lambda(4\hat{i} - \hat{j} + 4\hat{k})$.
Use the condition $\vec{c} \cdot \vec{d} = 16$, with $\vec{c} = \hat{i} - 2\hat{k}$:
\[ (\hat{i} + 0\hat{j} - 2\hat{k}) \cdot (\lambda(4\hat{i} - \hat{j} + 4\hat{k})) = 16 \]\[ \lambda [ (1)(4) + (0)(-1) + (-2)(4) ] = 16 \]\[ \lambda [ 4 - 8 ] = 16 \]\[ -4\lambda = 16 \implies \lambda = -4 \]Determine $\vec{d}$:
\[ \vec{d} = -4(4\hat{i} - \hat{j} + 4\hat{k}) = -16\hat{i} + 4\hat{j} - 16\hat{k} \]Calculate the magnitude of $\vec{d}$:
\[ |\vec{d}| = \sqrt{(-16)^2 + 4^2 + (-16)^2} = \sqrt{256 + 16 + 256} = \sqrt{528} \]Simplify $\sqrt{528}$: $528 = 16 \times 33$.
\[ |\vec{d}| = \sqrt{16 \times 33} = 4\sqrt{33} \]Alternative calculation: $|\vec{d}| = |\lambda| |\vec{a} \times \vec{b}| = |-4| |4\hat{i} - \hat{j} + 4\hat{k}| = 4\sqrt{4^2+(-1)^2+4^2} = 4\sqrt{16+1+16} = 4\sqrt{33}$.
Step 4: Conclusive Result:
The magnitude of $\vec{d}$ is $4\sqrt{33$}.