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

The correct answer is:
$$4\sqrt{33}$$

Step-by-Step Explanation:

We are given three vectors:
$\overrightarrow{a}=\hat{i}+4\hat{j}$
$\overrightarrow{b}=4\hat{j}+\hat{k}$
$\overrightarrow{c}=\hat{i}-2\hat{k}$

Step 1: Finding a vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$
Since the vector $\overrightarrow{d}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$, it must be parallel to their cross product, $\overrightarrow{a}\times \stackrel{��}{b}$.
Thus, we can write:
$$\overrightarrow{d}=\lambda (\overrightarrow{a}\times \overrightarrow{b})$$
where $\lambda$ is a scalar constant.

Step 2: Calculating the cross product $\overrightarrow{a}\times \overrightarrow{b}$
Using the determinant method for the cross product:
$$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 4& 0\\ 0& 4& 1\end{array}\right|$$
Expanding the determinant along the first row:
$$\overrightarrow{a}\times \overrightarrow{b}=\hat{i}(4·1-0·4)-\hat{j}(1·1-0·0)+\hat{k}(1·4-4·0)$$
$$\overrightarrow{a}\times \overrightarrow{b}=4\hat{i}-\hat{j}+4\hat{k}$$

Therefore, we can represent $\overrightarrow{d}$ as:
$$\overrightarrow{d}=\lambda (4\hat{i}-\hat{j}+4\hat{k})$$

Step 3: Determining $\lambda$ using the given dot product condition
We are given:
$$\overrightarrow{c}·\overrightarrow{d}=16$$
Substitute the values of $\overrightarrow{c}$ and $\overrightarrow{d}$:
$$(\hat{i}-2\hat{k})·\left[\lambda \right(4\hat{i}-\hat{j}+4\hat{k}\left)\right]=16$$
$$\lambda \left[\right(1\left)\right(4)+(0\left)\right(-1)+(-2\left)\right(4\left)\right]=16$$
$$\lambda [4-8]=16$$
$$-4\lambda =16$$
$$\lambda =-4$$

Step 4: Finding the magnitude $\left|\overrightarrow{d}\right|$
Now we write the vector $\overrightarrow{d}$:
$$\overrightarrow{d}=-4(4\hat{i}-\hat{j}+4\hat{k})$$
The magnitude is given by:
$$\left|\overrightarrow{d}\right|=|-4|\sqrt{4^2+{(-1)}^2+4^2}$$

$$\left|\overrightarrow{d}\right|=4\sqrt{33}$$