Find all vectors that are perpendicular to both vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + 2\hat{j} - 4\hat{k}$.

1. Problem Description

Find all vectors that are perpendicular to both vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + 2\hat{j} - 4\hat{k}$.

2. Solution Steps

A vector perpendicular to both $\vec{a}$ and $\vec{b}$ can be found by taking the cross product of the two vectors: $\vec{a} \times \vec{b}$. Any scalar multiple of this vector will also be perpendicular to both $\vec{a}$ and $\vec{b}$.

We compute the cross product as follows:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -2 & 2 & -4 \end{vmatrix}$

$= \hat{i} \begin{vmatrix} 2 & 3 \\ 2 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ -2 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -2 & 2 \end{vmatrix}$

$= \hat{i} (2(-4) - 3(2)) - \hat{j} (1(-4) - 3(-2)) + \hat{k} (1(2) - 2(-2))$

$= \hat{i} (-8 - 6) - \hat{j} (-4 + 6) + \hat{k} (2 + 4)$

$= -14\hat{i} - 2\hat{j} + 6\hat{k}$

Any vector of the form $c(-14\hat{i} - 2\hat{j} + 6\hat{k})$, where $c$ is a scalar, is perpendicular to both $\vec{a}$ and $\vec{b}$. We can simplify this by dividing by $-2$ to get $c(7\hat{i} + \hat{j} - 3\hat{k})$. So any vector of the form $k(7\hat{i} + \hat{j} - 3\hat{k})$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

3. Final Answer

$k(7\hat{i} + \hat{j} - 3\hat{k})$, where $k$ is a scalar.