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.

\\(vec{a} = ?\\) \\(vec{c} = ?\\)

Explanation:

Step1: Recall Right-Hand Rule

For cross - product \(\vec{A}\times\vec{B}=\vec{C}\), use right - hand rule: curl fingers from \(\vec{A}\) to \(\vec{B}\), thumb points to \(\vec{C}\) (out: \(\odot\), in: \(\otimes\)).

First Case (\(\vec{C}=\odot\) (out), \(\vec{B}\) right):

- Step1: Apply Right - Hand Rule
We know \(\vec{A}\times\vec{B}=\vec{C}\). Let's assume \(\vec{B}\) is along +x (right), \(\vec{C}\) is out of the page (\(\odot\)). By right - hand rule, if we curl from \(\vec{A}\) to \(\vec{B}\) (right), and thumb is out, then \(\vec{A}\) should be up (along +y direction).

Second Case (\(\vec{A}\) right, \(\vec{B}\) right? Wait, no, in the second diagram, \(\vec{B}\) is \(\otimes\) (in), \(\vec{A}\) is right.

- Step2: Apply Right - Hand Rule for \(\vec{C}\)
\(\vec{A}\) is right (+x), \(\vec{B}\) is in (-z). \(\vec{A}\times\vec{B}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\A_x&0&0\\0&0& - B_z\end{vmatrix}=\hat{j}(A_x(-B_z)-0)=\ - A_xB_z\hat{j}\), so \(\vec{C}\) is down ( - y direction).

Wait, let's re - analyze the first part:

First diagram: \(\vec{B}\) is right (let's say \(\vec{B}=B\hat{i}\)), \(\vec{C}=\odot\) (out, \(\vec{C}=C\hat{k}\)). From \(\vec{A}\times\vec{B}=\vec{C}\), let \(\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\). Then \(\vec{A}\times\vec{B}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\A_x&A_y&A_z\\B&0&0\end{vmatrix}=\hat{j}(0 - A_zB)-\hat{k}(0 - A_yB)=-A_zB\hat{j}+A_yB\hat{k}\). Since \(\vec{C}=C\hat{k}\), then \(A_z = 0\) and \(A_yB = C\). So for the direction of \(\vec{A}\), since the \(\hat{k}\) component comes from \(A_y\) (because the cross - product's \(\hat{k}\) term is \(A_yB\)), using right - hand rule: fingers from \(\vec{A}\) to \(\vec{B}\) (right), thumb to \(\vec{C}\) (out). So if we point right hand fingers along \(\vec{A}\), then curl towards \(\vec{B}\) (right), thumb is out. So \(\vec{A}\) must be up (along +y).

Second diagram: \(\vec{A}\) is right (\(\hat{i}\)), \(\vec{B}\) is \(\otimes\) (in, \(-\hat{k}\)). \(\vec{A}\times\vec{B}=\hat{i}\times(-\hat{k})=\hat{j}\)? Wait no, \(\hat{i}\times\hat{k}=-\hat{j}\), so \(\hat{i}\times(-\hat{k})=\hat{j}\)? Wait no, cross - product: \(\hat{i}\times\hat{j}=\hat{k}\), \(\hat{j}\times\hat{k}=\hat{i}\), \(\hat{k}\times\hat{i}=\hat{j}\), and anti - commutative: \(\hat{i}\times\hat{k}=-\hat{j}\). So \(\vec{A}=\hat{i}\) (right), \(\vec{B}=-\hat{k}\) (in), then \(\vec{A}\times\vec{B}=\hat{i}\times(-\hat{k})=\hat{j}\)? Wait, no, I think I messed up the direction of \(\vec{B}\) in the second diagram. The second diagram has \(\vec{B}\) as \(\otimes\) (in, so \(\vec{B}\) is into the page, \(-\hat{k}\)), \(\vec{A}\) is right (\(\hat{i}\)). Then \(\vec{A}\times\vec{B}=\hat{i}\times(-\hat{k})=\hat{j}\) (up)? Wait, no, let's use right - hand rule: hold your right hand, fingers along \(\vec{A}\) (right), then curl towards \(\vec{B}\) (into the page). To curl from right (x - direction) to into the page ( - z direction), your fingers will curl down? Wait, no, the right - hand rule for cross - product: the direction of \(\vec{A}\times\vec{B}\) is given by right hand: fingers along \(\vec{A}\), then bend fingers towards \(\vec{B}\), thumb points to \(\vec{C}\). So if \(\vec{A}\) is right (\(\hat{i}\)), \(\vec{B}\) is in (\(-\hat{k}\)), then to bend from \(\hat{i}\) to \(-\hat{k}\), your hand will rotate such that the thumb points down (\(-\hat{j}\))? Wait, maybe my coordinate system is wrong. Let's define:

- Right: \(\hat{i}\) (x - axis)
- Up: \(\hat{j}\) (y - axis)
- Out: \(\hat{k}\) (z - axis)
- Left: \(-\hat{i}\), Down: \(-\hat{j}\), In: \(-\hat{k}\)

First case: \(\vec{C}=\hat{k}\) (out), \(\vec{B}=\hat{i}\) (right). \(\vec{A}\times\vec{B}=\vec{C}\). So \(\vec{A}\times\hat{i}=\hat{k}\). We know that \(\hat{j}\times\hat{i}=-\hat{k}\), \(\hat{k}\times\hat{i}=\hat{j}\), \(\hat{i}\times\hat{i}=0\). Wait, no: \(\vec{A}\times\vec{B}=\vec{C}\), so \(\vec{A}\) must be such that when crossed with \(\vec{B}\) (right), gives out (up? No, out is z - axis). Wait, \(\hat{j}\times\hat{i}=-\hat{k}\), \(\hat{-j}\times\hat{i}=\hat{k}\). So \(\vec{A}=-\hat{j}\)? No, that can't be. Wait, maybe I got the direction of \(\vec{C}\) wrong. \(\odot\) is out (positive z), \(\otimes\) is in (negative z).

So \(\vec{A}\times\vec{B}=\vec{C}\). Let's solve for \(\vec{A}\)'s direction.

First problem: \(\vec{B}\) is right (\(\hat{i}\)), \(\vec{C}\) is out (\(\hat{k}\)). So \(\vec{A}\times\hat{i}=\hat{k}\). What vector \(\vec{A}\) when crossed with \(\hat{i}\) gives \(\hat{k}\)? Let's recall cross - product table:

\(\hat{i}\times\hat{j}=\hat{k}\), \(\hat{j}\times\hat{k}=\hat{i}\), \(\hat{k}\times\hat{i}=\hat{j}\)

\(\hat{j}\times\hat{i}=-\hat{k}\), \(\hat{k}\times\hat{j}=-\hat{i}\), \(\hat{i}\times\hat{k}=-\hat{j}\)

So we need \(\vec{A}\times\hat{i}=\hat{k}\). From the table, \(\hat{j}\times\hat{i}=-\hat{k}\), so \(-\hat{j}\times\hat{i}=\hat{k}\). So \(\vec{A}=-\hat{j}\)? No, that would be down. Wait, no, \(\hat{i}\times\vec{A}=-\vec{C}\)? Wait, cross - product is anti - commutative: \(\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}\). So if \(\vec{A}\times\vec{B}=\vec{C}\), then \(\vec{B}\times\vec{A}=-\vec{C}\).

In first case, \(\vec{B}=\hat{i}\), \(\vec{C}=\hat{k}\), so \(\vec{B}\times\vec{A}=-\hat{k}\). We know \(\hat{i}\times\vec{A}=-\hat{k}\). From cross - product table, \(\hat{i}\times\hat{j}=\hat{k}\), so \(\hat{i}\times(-\hat{j})=-\hat{k}\). So \(\vec{A}=-\hat{j}\)? No, that's down. But when we use right - hand rule: fingers along \(\vec{A}\) (down, \(-\hat{j}\)), curl towards \(\vec{B}\) (right, \(\hat{i}\)), thumb points out (\(\hat{k}\))? Let's try: right hand, fingers down (along \(-\hat{j}\)), then curl towards right (\(\hat{i}\)). The thumb will point out (\(\hat{k}\)). Yes! So \(\vec{A}\) is down? Wait, no, I'm confused.

Wait, let's start over with right - hand rule for \(\vec{A}\times\vec{B}=\vec{C}\):

1. Point your right hand fingers in the direction of \(\vec{A}\).
2. Curl your fingers towards \(\vec{B}\) (through the smallest angle between them).
3. Your thumb points in the direction of \(\vec{C}\).

First diagram:

- \(\vec{C}\) is out (thumb out), \(\vec{B}\) is right (curl towards right). So to have thumb out when curling from \(\vec{A}\) to \(\vec{B}\) (right), \(\vec{A}\) must be down. Wait, no: if \(\vec{A}\) is down (fingers down), curl towards right ( \(\vec{B}\) ), thumb is out. Yes! So \(\vec{A}\) is down? Wait, no, let's take an example: \(\vec{A}\) is down (\(-\hat{j}\)), \(\vec{B}\) is right (\(\hat{i}\)). Then \(\vec{A}\times\vec{B}=(-\hat{j})\times\hat{i}=\hat{k}\) (out), which matches \(\vec{C}=\hat{k}\). So \(\vec{A}\) is down? Wait, no, \((-\hat{j})\times\hat{i}=\hat{k}\), yes. So first \(\vec{A}\) direction: down? Wait, I think I made a mistake earlier.

Second diagram:

- \(\vec{A}\) is right (\(\hat{i}\)), \(\vec{B}\) is in (\(-\hat{k}\)). So fingers along \(\vec{A}\) (right), curl towards \(\vec{B}\) (in, \(-\hat{k}\)). The smallest angle between right (\(\hat{i}\)) and in (\(-\hat{k}\)) is 90 degrees (since \(\hat{i}\) and \(\hat{k}\) are perpendicular). So curling from \(\hat{i}\) to \(-\hat{k}\), the thumb points down (\(-\hat{j}\))? Wait, no: \(\hat{i}\times(-\hat{k})=\hat{j}\) (up). Wait, \(\hat{i}\times\hat{k}=-\hat{j}\), so \(\hat{i}\times(-\hat{k})=\hat{j}\). So thumb points up. But in the second diagram, \(\vec{C}\) is? Wait, the second diagram has \(\vec{A}\) right, \(\vec{B}\) in, find \(\vec{C}\). So \(\vec{A}\times\vec{B}=\hat{i}\times(-\hat{k})=\hat{j}\) (up). So \(\vec{C}\) is up.

Wait, I think I messed up the first analysis. Let's redo:

First problem: Find \(\vec{A}\) given \(\vec{A}\times\vec{B}=\vec{C}\), \(\vec{B}\) right, \(\vec{C}\) out.

Using right - hand rule:

- Thumb: \(\vec{C}\) (out)
- Curl: from \(\vec{A}\) to \(\vec{B}\) (right)
- So fingers: \(\vec{A}\) must be down (because when you curl from down to right, thumb is out).

\(\vec{A}\times\vec{B}\): \(\vec{A}\) down (\(-\hat{j}\)), \(\vec{B}\) right (\(\hat{i}\)): \((-\hat{j})\times\hat{i}=\hat{k}\) (out), correct.

Second problem: Find \(\vec{C}\) given \(\vec{A}\) right, \(\vec{B}\) in.

- Fingers: \(\vec{A}\) right, curl towards \(\vec{B}\) in ( \(-\hat{k}\) )
- Thumb: up (\(\hat{j}\)), so \(\vec{C}\) is up.

Wait, now I'm really confused. Let's use the cross - product formula \(\vec{A}\times\vec{B}=(A_yB_z - A_zB_y)\hat{i}+(A_zB_x - A_xB_z)\hat{j}+(A_xB_y - A_yB_x)\hat{k}\).

First case:

Assume \(\vec{B}=B\hat{i}\) (right), \(\vec{C}=C\hat{k}\) (out), \(\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\).

Then:

\(A_yB_z - A_zB_y = 0\) (since \(\vec{C}\) has no \(\hat{i}\) component)

\(A_zB_x - A_xB_z = 0\) (since \(\vec{C}\) has no \(\hat{j}\) component)

\(A_xB_y - A_yB_x = C\) ( \(\hat{k}\) component)

Since \(\vec{B}=B\hat{i}\), \(B_y = 0\), \(B_z = 0\), \(B_x = B\).

So:

First equation: \(A_y\times0 - A_z\times0 = 0\) (always true)

Second equation: \(A_z\times B - A_x\times0 = 0\Rightarrow A_z = 0\)

Third equation: \(A_x\times0 - A_y\times B = C\Rightarrow - A_yB = C\)

Since \(C>0\) (out), \(A_y=-\frac{C}{B}\), so \(\vec{A}\) has \(A_y<0\), i.e., down ( - y direction). So \(\vec{A}\) is down.

Second case:

\(\vec{A}=A\hat{i}\) (right), \(\vec{B}=B(-\hat{k})\) (in), find \(\vec{C}=\vec{A}\times\vec{B}\).

\(\vec{A}\times\vec{B}=A\hat{i}\times(-B\hat{k})=AB\hat{j}\) (up, since \(\hat{i}\times\hat{k}=-\hat{j}\), so \(\hat{i}\times(-\hat{k})=\hat{j}\)). So \(\vec{C}\) is up.

Answer:

First \(\vec{A}\) direction: down; Second \(\vec{C}\) direction: up