Question

1. (3pts each) given the cross - product \\( \vec { a } \times \vec { b } = \vec { c } \\), find the direction (left/right, in/out, up/down) of the missing vector in each case.

\\( a = ? \\)
\\( c = ? \\)

Explanation:

First Case (Top Diagram: $\vec{A}$, $\vec{B}$, $\vec{C}$)

Step1: Recall Cross-Product Rule

The cross - product $\vec{A}\times\vec{B}=\vec{C}$ follows the right - hand rule. For the right - hand rule, if we curl the fingers of our right hand from $\vec{A}$ to $\vec{B}$, the thumb points in the direction of $\vec{C}$. We know that $\vec{C}$ is out of the page (the $\odot$ symbol means out of the page) and $\vec{B}$ is to the right. Let's assume the direction of $\vec{A}$: let's consider the standard coordinate system where the positive x - direction is right, positive y - direction is up, and positive z - direction is out of the page. The cross - product formula in component form is $\vec{A}\times\vec{B}=(A_yB_z - A_zB_y,A_zB_x - A_xB_z,A_xB_y - A_yB_x)$. But using the right - hand rule: if $\vec{B}$ is along the positive x - axis ($\vec{B}=B\hat{i}$) and $\vec{C}$ is along the positive z - axis ($\vec{C}=C\hat{k}$), then from $\vec{A}\times\vec{B}=\vec{C}$, let $\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$. Then $\vec{A}\times\vec{B}=

$$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\A_x&A_y&A_z\\B&0&0\end{vmatrix}$$

=\hat{i}(0 - 0)-\hat{j}(0 - A_zB)+\hat{k}(0 - A_yB)=A_zB\hat{j}-A_yB\hat{k}$. For this to be equal to $C\hat{k}$, we need $A_zB = 0$ and $-A_yB=C$. Since $B
eq0$ (it has a direction), $A_z = 0$ and $A_y=-\frac{C}{B}$. But in terms of direction, using the right - hand rule: if we want $\vec{C}$ out of the page (thumb out), and $\vec{B}$ is to the right (fingers from $\vec{A}$ to $\vec{B}$), then $\vec{A}$ must be downward. Wait, let's do the right - hand rule properly: hold your right hand so that your fingers point in the direction of $\vec{A}$, then curl them towards $\vec{B}$ (right). The thumb should point in the direction of $\vec{C}$ (out). So if $\vec{B}$ is right, and $\vec{C}$ is out, then $\vec{A}$ must be down. Wait, no: let's take the standard right - hand rule for cross - product: $\hat{j}\times\hat{i}=-\hat{k}$, $\hat{k}\times\hat{i}=\hat{j}$, $\hat{i}\times\hat{j}=\hat{k}$. Wait, maybe I mixed up. Let's use the mnemonic: "Right hand, fingers from $\vec{A}$ to $\vec{B}$, thumb is $\vec{C}$". So if $\vec{C}$ is out (thumb out), and $\vec{B}$ is to the right (let's say $\vec{B}=\hat{i}$), then to get $\vec{C}=\hat{k}$, we need $\vec{A}=\hat{j}$ (up)? Wait, no: $\hat{j}\times\hat{i}=-\hat{k}$, $\hat{k}\times\hat{i}=\hat{j}$, $\hat{i}\times\hat{j}=\hat{k}$. Wait, $\vec{A}\times\vec{B}=\vec{C}$. So if $\vec{B}$ is $\hat{i}$ (right), and $\vec{C}$ is $\hat{k}$ (out), then $\vec{A}$ must be $\hat{j}$ (up)? Wait, no: $\hat{j}\times\hat{i}=-\hat{k}$, $\hat{i}\times\hat{j}=\hat{k}$. Oh! Right, $\vec{A}\times\vec{B}=\vec{C}$, so if $\vec{B}$ is $\hat{i}$ (right), $\vec{C}$ is $\hat{k}$ (out), then $\vec{A}$ must be $\hat{j}$ (up)? Wait, no, $\hat{i}\times\hat{j}=\hat{k}$, so $\vec{A}$ is $\hat{i}$? No, no: the order is $\vec{A}\times\vec{B}$, so the first vector is $\vec{A}$, second is $\vec{B}$. So if $\vec{A}$ is $\hat{j}$ (up) and $\vec{B}$ is $\hat{i}$ (right), then $\vec{A}\times\vec{B}=\hat{j}\times\hat{i}=-\hat{k}$ (into the page), which is not our case. If $\vec{A}$ is $\hat{-j}$ (down) and $\vec{B}$ is $\hat{i}$ (right), then $\vec{A}\times\vec{B}=(-\hat{j})\times\hat{i}=\hat{k}$ (out of the page), which matches $\vec{C}$. So $\vec{A}$ is downward.

Step2: Determine the Direction of $\vec{A}$

From the right - hand rule, when $\vec{C}$ is out of the page ( $\odot$ ) and $\vec{B}$ is to the right, the direction of $\vec{A}$ must be down.

Second Case (Bottom Diagram: $\vec{A}$, $\vec{B}$, $\vec{C}$)

Step1: Recall Cross - Product Rule

We know that $\vec{A}\times\vec{B}=\vec{C}$. $\vec{B}$ is to the right (let's say $\vec{B}=\hat{i}$), $\vec{A}$ is to the right? Wait, no, the diagram shows $\vec{A}$ and $\vec{B}$ as parallel? Wait, no, the bottom diagram: $\vec{A}$ is a horizontal arrow (let's assume right), $\vec{B}$ is a horizontal arrow (right), and $\vec{C}$ is? Wait, no, the symbol for $\vec{B}$ is a cross ( $\otimes$ ), which means into the page. Wait, the bottom diagram: $\vec{A}$ is a horizontal arrow (direction? Let's see, the arrow for $\vec{A}$ is to the right? Wait, no, the diagram: $\vec{A}$ has an arrow, $\vec{B}$ has a cross (into the page), and we need to find $\vec{C}$. Wait, using the right - hand rule: $\vec{A}\times\vec{B}=\vec{C}$. Let's assume $\vec{A}$ is to the right ( $\hat{i}$ ) and $\vec{B}$ is into the page ( $\hat{-k}$ ). Then $\vec{A}\times\vec{B}=\hat{i}\times(-\hat{k})=\hat{j}$ (up). Wait, let's do the right - hand rule: hold your right hand, fingers from $\vec{A}$ (right) to $\vec{B}$ (into the page). Curl your fingers from right to into the page. The thumb will point up. So $\vec{C}$ is up.

Answer:

s:

• For the top diagram (finding $\vec{A}$): The direction of $\vec{A}$ is down.
• For the bottom diagram (finding $\vec{C}$): The direction of $\vec{C}$ is up.