Why Does the Cross Product of î and ĵ Equal k̂?

[AakashPandita]

how is î x jcap = kcap? Please help!

 

[Fredrik]

Do you know the definition of the cross product?

 

[AakashPandita]

yes. a x b = absinθ

 

[pwsnafu]

yes. a x b = absinθ

Do you realize that the RHS of what you wrote is a scalar?

 

[Fredrik]

yes. a x b = absinθ

That's not the definition, and that equality isn't correct. You may be thinking of the result $\left|\mathbf a\times\mathbf b\right|=|\mathbf a||\mathbf b|\sin\theta$, where $\theta$ is the angle between the two vectors.

There are many equivalent ways to define the cross product. One of them is
$$(a_1,a_2,a_3)\times(b_1,b_2,b_3)=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1).$$ You should check what definition your book uses, and then try to use it to prove that it implies that
$$\mathbf i\times\mathbf j=\mathbf k.$$

 

[pwsnafu]

That's not the definition, and that equality isn't correct. You may be thinking of the result $\mathbf a\times\mathbf b=|\mathbf a||\mathbf b|\sin\theta$, where $\theta$ is the angle between the two vectors.

Again LHS is a vector, RHS is a scalar.

 

[Fredrik]

Again LHS is a vector, RHS is a scalar.

LOL, yes I know. That's why I started typing that. Somehow I forgot to type the absolute value symbols on the left. I will edit my post.

 

[AakashPandita]

Oh I understood. (ab sin theta) was only the magnitude.

Thank you very very much!

 

[HallsofIvy]

One good way of defining the cross product is to start with
1)\vec{i}\times\vec{j}= \vec{k}
2)\vec{j}\times\vec{k}=\vec{i}
3)\vec{k}\times\vec{i}= \vec{j}
Then extend it to all other vectors by "linearity" in the first component:
(\vec{u}+ \vec{v})\times \vec{w}= \vec{u}\times \vec{w}+ \vec{v}\times \vec{w}
and by "anti- commutativity":
\vec{u}\times\vec{v}= -\vec{v}\times\vec{u}

What you are asking about is (1) above.

Another, equivalent but less "sophisticated", way to define the cross product is to simply say that
(A\vec{i}+ B\vec{j}+ C\vec{k})\times(P\vec{i}+ Q\vec{j}+ R\vec{k})= (BR- CQ)\vec{i}- (AR- CP)\vec{j}+ (AQ- BP)\vec{k}

and then, \vec{i}\times\vec{j} has A=1, B= 0, C= 0, P= 0, Q= 1, R= 0
so the product is (0(0)- 0(1))\vec{i}- (1(0)- 0(0))\vec{j}+ (1(1)- 0(0))\vec{k}= \vec{k}