Vector algebra in component form

Now, how do we add two vectors in this component form?

If we have vector A and vector B we want to add them, A has x, y, and z components: Ax, Ay, and Az. B has Bx, By, and Bz.

So, this in the component form can be written as (Ax plus Bx) x cap plus (Ay plus By) y cap plus (Az plus Bz) z cap.

How do we multiply a scalar with a vector written in its component form? A scalar a is multiplied with a vector Ax x cap plus Ay y cap plus Az z cap, which just gives a this vector.

So, we have in the Cartesian coordinate system x cap, y cap, and z cap; they are mutually perpendicular. These are orthogonal vectors, and the magnitude of all these unit vectors is 1.

Therefore, we have x cap dot x cap equals y cap dot y cap, which is the same equals z cap dot z cap is 1.

If we take the dot product of two different unit vectors, that is x cap dot y cap, that equals x cap dot z cap, which equals y cap dot z cap; that always becomes 0.

In a more compact notation, we can write i cap dot j cap equals Kronecker delta ij.

Now, i and j can take values from x, y, and z. So, we can write i, j belongs to the set x, y, z.

The Kronecker delta function is defined as 1 if i equals j and 0 if i does not equal j; that is the definition of Kronecker delta.

Accordingly, now if we make a dot product of A and B; so, A dot B, where A and B are written in component form, is written as Ax x cap plus Ay y cap plus Az z cap dot Bx x cap plus By y cap plus Bz z cap.

Now, if we take a dot product between this part and this part, we obtain Ax Bx.

If we take a dot product between this part and this part, because the dot product of x cap and y cap goes to 0, we get 0 here.

For the next one also, we get 0.

So, the dot product between Ax x cap and Bz z cap also becomes 0.

Then we come to Ay y cap; its dot product with Bx x cap goes to 0.

The dot product that sustains is By y cap.

So, for that, we get Ay By, and the dot product with Bz z cap for Ay y cap also goes to 0.

Similarly, for Az z cap, we will have only the dot product sustaining with Bz z cap.

So, we get here Az Bz; that means, the dot product that we are interested in becomes Ax Bx plus Ay By plus Az Bz.

Now, if we take a dot product of vector A with itself, A dot A, what do we get as a result?

Looking at the example earlier, we can write that it would become Ax squared plus Ay squared plus Az squared.

If that is the case, then we can write that the magnitude of vector A is actually the square root of this part, Ax squared plus Ay squared plus Az squared.

Now let us go for the cross product in this component form because with the cross product we get a vector perpendicular to the plane.

Therefore, x cap cross x cap becomes 0.

Similarly, y cap cross y cap and z cap cross z cap are also 0.

If we take x cap cross y cap, that is minus y cap cross x cap, as we have found out for any two vectors earlier, and this equals z cap in the orthogonal Cartesian coordinate system.

Similarly, y cap cross z cap equals minus z cap cross y cap equals x cap, and z cap cross x cap equals minus x cap cross z cap equals y cap.

This will be useful later for performing any cross product.

So, if we now perform a cross product between two vectors A and B, we can write it as Ax x cap plus Ay y cap plus Az z cap cross Bx x cap plus By y cap plus Bz z cap.

That becomes, going by the rules that we have set here, (Ay Bz minus Az By) x cap plus (Az Bx minus Ax Bz) y cap plus (Ax By minus Ay Bx) z cap.

Now, this expression, the way we have performed the cross product, is by going through the relations we have set here above, but there is a better way to remember it and represent it in a compact form, that is by making a determinant.

So, the cross product can be written in a determinant form as follows: A cross B with its Cartesian components can be written as a determinant x cap y cap z cap; Ax Ay Az; Bx By Bz.

Let us see an example.

Let us consider a cube of arm length one in the Cartesian coordinate system.

Here is our Cartesian coordinate system; these are the axes x, y, and z.

So, this much is arm length one.

We have the cube like this; here is the cube.

And we have two vectors; this is vector A and here is vector B.

We want to find the angle between these two vectors; that means, this angle, let us call it theta.

So, let us write down A in its coordinate form.

If we look at it, we can see that A projected onto the x-axis will give us 1; that means, its x component is 1, 1 x cap.

Projected onto the y-axis, there is no projection; it is 0.

If we project it onto the z-axis, that is also 1.

So, this is the expression; this is the component form for vector A.

For vector B, the component form would look like its projection onto the x-axis is 0.

So, 0 x cap plus projection onto y is 1, so 1 y cap, and projection onto z is also 1.

So, 1 z cap; this is the form of vector B.

Now, if we perform a dot product of A and B using the component forms that we have just deduced, we have this: A dot B equals 1 dot 0 plus 0 dot 1 plus 1 dot 1; that means, it is 1.

We know that A dot B equals A B cosine of the angle between them.

So, let us find out the magnitude of vector A; that is given as 1 squared plus 0 squared plus 1 squared, square root of this entire thing, that is square root of 2.

For vector B, the magnitude is also the same: 0 squared plus 1 squared plus 1 squared, square root of this equals square root of 2.

Now, applying this equation here, we find that square root 2 times square root 2 times cosine of the angle, that is 2 cos theta equals 1.

If 2 cos theta equals 1; that means, cos theta equals half, which gives us theta equals 60 degrees.