Let $\vec a = 2\hat i-\hat j, \vec b = -4\hat j+\hat k$ 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.\vec d=34, then $|\vec d|$ is equal to

$2\sqrt{69}$

The correct answer is Option (2) → $2\sqrt{69}$

Given:

$\vec{a} = 2\hat{i} - \hat{j}$,   $\vec{b} = -4\hat{j} + \hat{k}$,   $\vec{c} = \hat{i} + 2\hat{k}$

$\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ ⇒ $\vec{d}$ is parallel to $\vec{a} \times \vec{b}$

So, $\vec{d} = \lambda (\vec{a} \times \vec{b})$ for some scalar $\lambda$

Step 1: Compute $\vec{a} \times \vec{b}$

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 0 \\ 0 & -4 & 1 \\ \end{vmatrix} = \hat{i}(-1 \cdot 1 - 0 \cdot (-4)) - \hat{j}(2 \cdot 1 - 0 \cdot 0) + \hat{k}(2 \cdot (-4) - 0 \cdot 0)$

$\Rightarrow \vec{a} \times \vec{b} = -\hat{i} - 2\hat{j} - 8\hat{k}$

Step 2: Let $\vec{d} = \lambda(-\hat{i} - 2\hat{j} - 8\hat{k})$

Now given: $\vec{c} \cdot \vec{d} = 34$

$\vec{c} \cdot \vec{d} = (\hat{i} + 2\hat{k}) \cdot \lambda(-\hat{i} - 2\hat{j} - 8\hat{k})$

$= \lambda[(1)(-1) + (0)(-2) + (2)(-8)] = \lambda(-1 - 16) = -17\lambda$

$-17\lambda = 34 \Rightarrow \lambda = -2$

Step 3: $\vec{d} = -2(-\hat{i} - 2\hat{j} - 8\hat{k}) = 2\hat{i} + 4\hat{j} + 16\hat{k}$

Magnitude: $|\vec{d}| = \sqrt{2^2 + 4^2 + 16^2} = \sqrt{4 + 16 + 256} = \sqrt{276}$

Final Answer: $|\vec{d}| = \sqrt{276} = 2\sqrt{69}$