The problem asks us to find the magnitude of a vector $\vec{d}$ that is perpendicular to two given vectors, $\vec{a} = \hat{i}+4\hat{j}$ and $\vec{b} = 4\hat{j} + \hat{k}$, and also satisfies the condition $\vec{c} \cdot \vec{d} = 16$, where $\vec{c} = \hat{i}-2\hat{k}$.

Vector Properties for Solution

A vector that is perpendicular to two vectors is parallel to their cross product. Therefore, the vector $\vec{d}$ must be parallel to $\vec{a} \times \vec{b}$. We can express $\vec{d}$ as:

$ \vec{d} = k (\vec{a} \times \vec{b}) $

where $k$ is a scalar constant.

Calculating the Cross Product $\vec{a} \times \vec{b}$

First, let's express the vectors $\vec{a}$ and $\vec{b}$ in component form:

• $\vec{a} = \hat{i}+4\hat{j} = 1\hat{i} + 4\hat{j} + 0\hat{k} = \langle 1, 4, 0 \rangle$
• $\vec{b} = 4\hat{j} + \hat{k} = 0\hat{i} + 4\hat{j} + 1\hat{k} = \langle 0, 4, 1 \rangle$

Now, we compute the cross product $\vec{a} \times \vec{b}$ using the determinant formula:

$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 0 \\ 0 & 4 & 1 \end{vmatrix} $

Expanding the determinant:

$ \vec{a} \times \vec{b} = \hat{i}(4 \cdot 1 - 0 \cdot 4) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 4 - 4 \cdot 0) $

$ \vec{a} \times \vec{b} = \hat{i}(4 - 0) - \hat{j}(1 - 0) + \hat{k}(4 - 0) $

$ \vec{a} \times \vec{b} = 4\hat{i} - \hat{j} + 4\hat{k} = \langle 4, -1, 4 \rangle $

Determining the Scalar Constant $k$

We know that $\vec{d} = k(4\hat{i} - \hat{j} + 4\hat{k})$. Now let's express the vector $\vec{c}$ in component form:

$ \vec{c} = \hat{i}-2\hat{k} = 1\hat{i} + 0\hat{j} - 2\hat{k} = \langle 1, 0, -2 \rangle $

We are given the condition $\vec{c} \cdot \vec{d} = 16$. Let's calculate the dot product:

$ \vec{c} \cdot \vec{d} = \langle 1, 0, -2 \rangle \cdot \langle 4k, -k, 4k \rangle $

$ \vec{c} \cdot \vec{d} = (1)(4k) + (0)(-k) + (-2)(4k) $

$ \vec{c} \cdot \vec{d} = 4k + 0 - 8k $

$ \vec{c} \cdot \vec{d} = -4k $

Using the given condition:

$ -4k = 16 $

Solving for $k$:

$ k = \frac{16}{-4} = -4 $

Calculating the Magnitude $|\vec{d}|$

Now that we have the scalar $k = -4$, we can find the magnitude of $\vec{d}$. We know:

$ \vec{d} = k (\vec{a} \times \vec{b}) $

The magnitude is given by:

$ |\vec{d}| = |k| |\vec{a} \times \vec{b}| $

We already calculated $\vec{a} \times \vec{b} = \langle 4, -1, 4 \rangle$. Its magnitude is:

$ |\vec{a} \times \vec{b}| = \sqrt{4^2 + (-1)^2 + 4^2} $

$ |\vec{a} \times \vec{b}| = \sqrt{16 + 1 + 16} $

$ |\vec{a} \times \vec{b}| = \sqrt{33} $

Now, substitute the values of $|k|$ and $|\vec{a} \times \vec{b}|$ into the equation for $|\vec{d}|$:

$ |\vec{d}| = |-4| \cdot \sqrt{33} $

$ |\vec{d}| = 4 \sqrt{33} $

Final Result

The magnitude of the vector $\vec{d}$ is $4\sqrt{33}$.