In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given , the vector form is $\displaystyle a=\left \langle a_{1},a_{2},a_{3}\right \rangle$.

So for $\displaystyle 3i-5j+7k$, we can derive the vector form $\displaystyle \left \langle 3,-5,7\right \rangle$.

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given , the vector form is .

So for $\displaystyle 7i$, we can derive the vector form $\displaystyle \left \langle 7,0,0\right \rangle$.

In order to derive the vector form of the distance between two points, we must find the difference between the , , and  elements of the points.

That is, for any point 

 and $\displaystyle b=\left \langle b_1,b_2,b_3\right \rangle$,

the distance is the vector 

$\displaystyle v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$. 

Subbing in our original points  and ,  we get:

$\displaystyle v=\left \langle 5-9,2-4,7-(-1)\right \rangle$

$\displaystyle v=\left \langle -4,-2,8\right \rangle$

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given, the vector form is  .

So for $\displaystyle -6j$, we can derive the vector form $\displaystyle \left \langle 0,-6,0 \right \rangle$.

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point

and ,

the distance is the vector

.

Subbing in our original points $\displaystyle (0,1,3)$ and $\displaystyle (5,7,4)$, we get:

$\displaystyle v=\left \langle 5-0,7-1,4-3\right \rangle$

$\displaystyle v=\left \langle 5,6,1\right \rangle$

In order to derive the vector form of the distance between two points, we must find the difference between the , , and  elements of the points.

That is, for any point 

 and ,

the distance is the vector 

.

Subbing in our original points $\displaystyle (2,4,5)$ and $\displaystyle (9,1,0)$, we get:

$\displaystyle v=\left \langle 9-2,1-4,0-5\right \rangle$

$\displaystyle v=\left \langle 7,-3,-5\right \rangle$

In order to derive the vector form of the distance between two points, we must find the difference between the , , and  elements of the points. That is, for any point  and , the distance is the vector .

Subbing in our original points $\displaystyle (0,0,-4)$ and $\displaystyle (4,9,-1)$, we get:

 $\displaystyle v=\left \langle 4-0,9-0,-1+4\right \rangle$

$\displaystyle v=\left \langle 4,9,3\right \rangle$

The derivative of a vector function has its components consisting of the derivatives of its components:

$\displaystyle r'(t) = (-\sin t,\cos t)$

since the derivatives of $\displaystyle \cos t, \sin t$ are $\displaystyle -\sin t, \cos t$, respectively.

The derivative of a vector function is just the derivative of components:

$\displaystyle r'(t)=(f_1'(t),f_2'(t),f_3'(t))=(1,2t,3t^2)$

The derivative of each component was found using the power rule which states,

$\displaystyle x^n dx=nx^{n-1}$.

Thus,

$\displaystyle t\ dt=1\cdot t^{1-1}=1\cdot t^0=1$,

$\displaystyle t^2dt=2t^{2-1}=2t^1=2t$,

$\displaystyle t^3dt=3t^{3-1}=3t^2$.

All of the vector forms above describe the circle $\displaystyle x^2+y^2=1$ since if we square their components and add them, we get (using trig. identities): 

$\displaystyle (x,y)\rightarrow (cos(x), sin(x))$

Therefore we get,

$\displaystyle x^2+y^2=\sin^2+\cos^2=1$.