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: Understanding the Concept:
A vector that is perpendicular to two other vectors, $\vec{a}$ and $\vec{b}$, must be parallel to their cross product, $\vec{a} \times \vec{b}$. We can express $\vec{d}$ as a scalar multiple of this cross product and then use the given dot product condition to find the scalar. Finally, we calculate the magnitude of $\vec{d}$.
Step 2: Key Formula or Approach:
1. Calculate the cross product $\vec{n} = \vec{a} \times \vec{b}$. 2. Let $\vec{d} = \lambda \vec{n}$ for some scalar $\lambda$. 3. Use the condition $\vec{c} \cdot \vec{d} = 16$ to solve for $\lambda$. 4. Calculate the magnitude $|\vec{d}|$.
Step 3: Detailed Explanation:
Given vectors are $\vec{a} = \hat{i} + 4\hat{j} + 0\hat{k}$ and $\vec{b} = 0\hat{i} + 4\hat{j} + \hat{k}$.
First, find 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 both $\vec{a}$ and $\vec{b}$, it must be parallel to their cross product.
Let $\vec{d} = \lambda(4\hat{i} - \hat{j} + 4\hat{k})$ for some scalar $\lambda$.
We are given that $\vec{c} \cdot \vec{d} = 16$, where $\vec{c} = \hat{i} - 2\hat{k}$.
\[ (\hat{i} + 0\hat{j} - 2\hat{k}) \cdot (\lambda(4\hat{i} - \hat{j} + 4\hat{k})) = 16 \] \[ \lambda [ (1)(4) + (0)(-1) + (-2)(4) ] = 16 \] \[ \lambda [ 4 + 0 - 8 ] = 16 \] \[ \lambda [ -4 ] = 16 \implies \lambda = -4 \] Now we have the vector $\vec{d}$:
\[ \vec{d} = -4(4\hat{i} - \hat{j} + 4\hat{k}) = -16\hat{i} + 4\hat{j} - 16\hat{k} \] Finally, find the magnitude of $\vec{d}$:
\[ |\vec{d}| = \sqrt{(-16)^2 + 4^2 + (-16)^2} = \sqrt{256 + 16 + 256} = \sqrt{528} \] To simplify $\sqrt{528}$: $528 = 16 \times 33$.
\[ |\vec{d}| = \sqrt{16 \times 33} = 4\sqrt{33} \] Alternatively, $|\vec{d}| = |\lambda| |\vec{a} \times \vec{b}| = |-4| |4\hat{i} - \hat{j} + 4\hat{k}| = 4\sqrt{4^2+(-1)^2+4^2} = 4\sqrt{16+1+16} = 4\sqrt{33}$.
Step 4: Final Answer:
The magnitude of $\vec{d}$ is $4\sqrt{33$}.