unit vectors

[logistic_guy]

Show that \(\displaystyle \bold{(a)} \ \bold{i} \times \bold{i} = \bold{j} \times \bold{j} = \bold{k} \times \bold{k} = 0\), \(\displaystyle \bold{(b)} \ \bold{i} \times \bold{j} = \bold{k}, \bold{i} \times \bold{k} = -\bold{j}, \text{and} \ \bold{j} \times \bold{k} = \bold{i}\).

 

[khansaheb]

Show that \(\displaystyle \bold{(a)} \ \bold{i} \times \bold{i} = \bold{j} \times \bold{j} = \bold{k} \times \bold{k} = 0\), \(\displaystyle \bold{(b)} \ \bold{i} \times \bold{j} = \bold{k}, \bold{i} \times \bold{k} = -\bold{j}, \text{and} \ \bold{j} \times \bold{k} = \bold{i}\).

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle \bold{(a)}\)

In the kindergarten, they taught us that the magnitude of the cross product between two vectors is:

\(\displaystyle |\bold{a} \times \bold{b}| = |\bold{a}||\bold{b}|\sin\theta\)

Then,

\(\displaystyle |\bold{i} \times \bold{i}| = |\bold{i}||\bold{i}|\sin 0 = 1 \times 1 \times 0 = 0\)
\(\displaystyle |\bold{j} \times \bold{j}| = |\bold{j}||\bold{j}|\sin 0 = 1 \times 1 \times 0 = 0\)
\(\displaystyle |\bold{k} \times \bold{k}| = |\bold{k}||\bold{k}|\sin 0 = 1 \times 1 \times 0 = 0\)

If the magnitude of a vector \(\displaystyle |\bold{a} \times \bold{b}| = 0\), then the vector itself \(\displaystyle \bold{a} \times \bold{b} = 0\) too.

Then,

\(\displaystyle \bold{i} \times \bold{i} = \bold{j} \times \bold{j} = \bold{k} \times \bold{k} = 0\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

If the cross product between two vectors gives a unit vector, then its magnitude must be \(\displaystyle 1\). Let us verify this fact first.

\(\displaystyle |\bold{i} \times \bold{j}| = |\bold{i}||\bold{j}|\sin 90^\circ = 1 \times 1 \times 1 = 1 = |\bold{k}|\)
\(\displaystyle |\bold{i} \times \bold{k}| = |\bold{i}||\bold{k}|\sin 90^\circ = 1 \times 1 \times 1 = 1 = |-\bold{j}|\)
\(\displaystyle |\bold{j} \times \bold{k}| = |\bold{j}||\bold{k}|\sin 90^\circ = 1 \times 1 \times 1 = 1 = |\bold{i}|\)

Now if we use the right hand rule with our middle finger pointing to the \(\displaystyle \bold{j}\) direction, we see that our thumb points in the \(\displaystyle \bold{k}\) direction, then \(\displaystyle \bold{i} \times \bold{j} = \bold{k}\).

Again, if we let our middle finger pointing to the \(\displaystyle \bold{k}\) direction, we see that our thumb points in the \(\displaystyle \bold{-j}\) direction, then \(\displaystyle \bold{i} \times \bold{k} = \bold{-j}\).

We repeat, if we let our middle finger pointing to the \(\displaystyle \bold{k}\) direction, we see that our thumb points in the \(\displaystyle \bold{i}\) direction, then \(\displaystyle \bold{j} \times \bold{k} = \bold{i}\).

\(\displaystyle \textcolor{red}{\bold{Note}}\) that in all of the three cases above we used our second finger (index finger) to point in the direction of the first unit vector.