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 \(\vec{C}\). Or, if \(\vec{C}\) is given (direction: \(\odot\) is out, \(\otimes\) is in), and \(\vec{B}\) is given, find \(\vec{A}\); or if \(\vec{A}\) and \(\vec{B}\) are given, find \(\vec{C}\).

First Case (\(\vec{A}\times\vec{B}=\vec{C}\), find \(\vec{A}\)'s direction? Wait, no, first diagram: \(\vec{B}\) is horizontal right, \(\vec{C}\) is out (\(\odot\)).

Using right - hand rule: Let \(\vec{A}\) be vertical (up/down). If \(\vec{B}\) is right, \(\vec{C}\) is out. Curl from \(\vec{A}\) to \(\vec{B}\), thumb out. So \(\vec{A}\) should be up (because if \(\vec{A}\) is up, \(\vec{B}\) is right, cross product \(\vec{A}\times\vec{B}\) is out (right - hand rule: up (y - axis) \(\times\) right (x - axis) = out (z - axis positive)).

Second Case: \(\vec{A}\) is horizontal right, \(\vec{B}\) is? Wait, second diagram: \(\vec{A}\) is horizontal right, \(\vec{B}\) is? Wait, \(\vec{C}\) is? Wait, the second part: \(\vec{A}\) is right, \(\vec{B}\) is? Wait, the diagram has \(\vec{A}\) right, \(\vec{B}\) is? Wait, the cross - product \(\vec{A}\times\vec{B}=\vec{C}\). \(\vec{A}\) is right (x - axis), \(\vec{B}\) is? Wait, the symbol for \(\vec{B}\) is \(\otimes\)? No, \(\vec{B}\)'s direction? Wait, no, in the second case, we have \(\vec{A}\) right, \(\vec{B}\) is? Wait, the problem is two sub - questions:

Sub - question 1: Find \(\vec{A}\) (direction) when \(\vec{B}\) is right, \(\vec{C}\) is out (\(\odot\))

Using right - hand rule: \(\vec{A}\times\vec{B}=\vec{C}\). Let \(\vec{A}\) be along y - axis, \(\vec{B}\) along x - axis. \(\hat{y}\times\hat{x}=-\hat{z}\)? Wait, no, wait: \(\hat{i}\) (x), \(\hat{j}\) (y), \(\hat{k}\) (z). \(\hat{j}\times\hat{i}=-\hat{k}\), \(\hat{i}\times\hat{j}=\hat{k}\), \(\hat{k}\times\hat{i}=\hat{j}\), etc. Wait, I think I messed up. Let's re - do:

Right - hand rule: For \(\vec{A}\times\vec{B}\), point right hand fingers in \(\vec{A}\)'s direction, curl towards \(\vec{B}\)'s direction, thumb is \(\vec{C}\)'s direction.

First diagram: \(\vec{C}\) is out (\(\odot\)) (thumb out), \(\vec{B}\) is right (fingers curl from \(\vec{A}\) to \(\vec{B}\)). So to get thumb out, when \(\vec{B}\) is right, \(\vec{A}\) must be up (because if \(\vec{A}\) is up, fingers from up to right, curl, thumb out).

Second diagram: \(\vec{A}\) is right, \(\vec{B}\) is? Wait, the symbol for \(\vec{B}\) is \(\otimes\)? No, the second diagram: \(\vec{A}\) is right, \(\vec{B}\) is? Wait, the cross - product \(\vec{A}\times\vec{B}=\vec{C}\). \(\vec{A}\) is right (\(\hat{i}\)), \(\vec{B}\) is? Wait, the \(\vec{C}\) is? Wait, the problem says "find the direction (left/right, in/out, up/down) of the missing vector in each case".

First case: Missing vector is \(\vec{A}\), \(\vec{B}\) is right, \(\vec{C}\) is out. So \(\vec{A}\) direction: up (because \(\vec{A}\times\vec{B}=\vec{C}\), \(\vec{A}\) up, \(\vec{B}\) right, \(\vec{C}\) out (by right - hand rule: up (y) \(\times\) right (x) = out (z)? Wait, no: \(\hat{y}\times\hat{x}=-\hat{z}\) (into the page), \(\hat{x}\times\hat{y}=\hat{z}\) (out). Wait, I had the order wrong. Cross - product is \(\vec{A}\times\vec{B}\), so the order is \(\vec{A}\) first, then \(\vec{B}\). So if \(\vec{C}=\vec{A}\times\vec{B}\) is out (\(\hat{z}\)), and \(\vec{B}\) is \(\hat{x}\) (right), then \(\vec{A}\) must be \(\hat{y}\) (up), because \(\hat{y}\times\hat{x}=-\hat{z}\) (no, that's into). Wait, \(\hat{x}\times\hat{y}=\hat{z}\) (out). So if \(\vec{C}\) is out (\(\hat{z}\)), and \(\vec{B}\) is \(\hat{y}\)? No, the diagram: \(\vec{B}\) is horizontal right (x - axis), \(\vec{C}\) is out (z - axis positive). So \(\vec{A}\) must be up (y - axis), because \(\vec{A}\) (y) \(\times\) \(\vec{B}\) (x) = - z (into), no. Wait, maybe I got the direction of \(\vec{B}\) wrong. Wait, the first diagram: \(\vec{B}\) is a horizontal line with arrow right, so \(\vec{B}\) is along positive x - axis. \(\vec{C}\) is \(\odot\), which is positive z - axis (out). So \(\vec{A}\times\vec{B}=\vec{C}\), so \(\vec{A}\times\hat{x}=\hat{z}\). We know that \(\hat{y}\times\hat{x}=-\hat{z}\), \(\hat{z}\times\hat{x}=-\hat{y}\), \(\hat{x}\times\hat{x}=0\), \(\hat{-y}\times\hat{x}=\hat{z}\). Ah! So \(\vec{A}\) is along negative y - axis (down). Because \((-\hat{y})\times\hat{x}=\hat{z}\) (since \(\hat{y}\times\hat{x}=-\hat{z}\), so \(-\hat{y}\times\hat{x}=\hat{z}\)). So that's the mistake earlier. So to get \(\vec{A}\times\vec{B}=\vec{C}\) (out), with \(\vec{B}\) right (x), \(\vec{A}\) must be down ( - y).

Second case: \(\vec{A}\) is right (x), \(\vec{B}\) is? Wait, the diagram has \(\vec{A}\) right, \(\vec{B}\) is a circle with cross (\(\otimes\)), which is into the page ( - z), and we need to find \(\vec{C}=\vec{A}\times\vec{B}\). \(\vec{A}\) is \(\hat{x}\), \(\vec{B}\) is \(-\hat{z}\) (into). Then \(\vec{A}\times\vec{B}=\hat{x}\times(-\hat{z})=\hat{y}\) (up). Wait, no: \(\hat{x}\times\hat{z}=-\hat{y}\), so \(\hat{x}\times(-\hat{z})=\hat{y}\) (up). Wait, the problem says "find the direction of the missing vector in each case".

First case: Missing vector is \(\vec{A}\), \(\vec{B}\) right, \(\vec{C}\) out. So \(\vec{A}\) direction: down (because \(\vec{A}\) (down, - y) \(\times\) \(\vec{B}\) (right, x) = out (z), since \(-\hat{y}\times\hat{x}=\hat{z}\)).

Second case: \(\vec{A}\) right, \(\vec{B}\) is? Wait, \(\vec{C}\) is? Wait, the second diagram: \(\vec{A}\) is right, \(\vec{B}\) is into the page (\(\otimes\)), find \(\vec{C}=\vec{A}\times\vec{B}\). Using right - hand rule: \(\vec{A}\) is right (x), \(\vec{B}\) is into ( - z). So \(\vec{A}\times\vec{B}=\hat{x}\times(-\hat{z})=\hat{y}\) (up). Wait, no, \(\hat{x}\times\hat{z}=-\hat{y}\), so \(\hat{x}\times(-\hat{z})=\hat{y}\) (up). So \(\vec{C}\) is up.

Wait, maybe the first part: \(\vec{B}\) is right, \(\vec{C}\) is out, find \(\vec{A}\)'s direction. Using right - hand rule correctly: For \(\vec{A}\times\vec{B}=\vec{C}\), hold your right hand so that your fingers point in the direction of \(\vec{A}\), then curl them towards \(\vec{B}\); your thumb points in the direction of \(\vec{C}\).

So if \(\vec{C}\) is out (thumb out), \(\vec{B}\) is right (fingers curl from \(\vec{A}\) to \(\vec{B}\) towards right). So to have thumb out, \(\vec{A}\) must be down (because if \(\vec{A}\) is down, fingers from down to right, curl, thumb out).

Second part: \(\vec{A}\) is right, \(\vec{B}\) is into the page (\(\otimes\)), find \(\vec{C}\). Fingers point right (\(\vec{A}\)), curl towards into the page (\(\vec{B}\)), thumb points up. So \(\vec{C}\) is up.

Step1 (First Sub - Question: Find \(\vec{A}\)'s direction when \(\vec{B}\) is right, \(\vec{C}\) is out)

Using right - hand rule for \(\vec{A}\times\vec{B}=\vec{C}\). Let \(\vec{B}=\hat{x}\) (right), \(\vec{C}=\hat{z}\) (out). We know that \(\vec{A}\times\hat{x}=\hat{z}\). From cross - product formula, \(\vec{A}=a\hat{y}+b\hat{z}+c\hat{x}\), but since \(\vec{B}\) is \(\hat{x}\), \(\vec{A}\times\hat{x}=(a\hat{y}+b\hat{z}+c\hat{x})\times\hat{x}=a(\hat{y}\times\hat{x})+b(\hat{z}\times\hat{x})+c(\hat{x}\times\hat{x})=-a\hat{z}-b\hat{y}+0\). Set equal to \(\hat{z}\), so \(-a\hat{z}-b\hat{y}=\hat{z}\). Thus, \(a = - 1\), \(b = 0\). So \(\vec{A}\) has component \(-\hat{y}\) (down). So \(\vec{A}\) is down.

Step2 (Second Sub - Question: Find \(\vec{C}\)'s direction when \(\vec{A}\) is right (\(\hat{x}\)) and \(\vec{B}\) is into (\(-\hat{z}\)))

\(\vec{C}=\vec{A}\times\vec{B}=\hat{x}\times(-\hat{z})\). Using cross - product rule \(\hat{x}\times\hat{z}=-\hat{y}\), so \(\hat{x}\times(-\hat{z})=\hat{y}\) (up). So \(\vec{C}\) is up.

Answer:

First case (missing \(\vec{A}\)): down
Second case (missing \(\vec{C}\)): up