The difference \(\displaystyle a-b\) of two vectors \(\displaystyle < a_1,a_2>\) and \(\displaystyle < b_1,b_2>\) is defined as

\(\displaystyle < a_1-b_1,a_2-b_2>\)

For the two vectors 

\(\displaystyle u=< 2,3>\)

\(\displaystyle v=< 0,1>\)

The difference is

\(\displaystyle u-v=< 2,3>-< 0,1>\)

\(\displaystyle =< 2-0,3-1>\)

\(\displaystyle =< 2,2>\)

Given the vectors

\(\displaystyle v=(1,3)\)

\(\displaystyle w=(5,2)\)

we can find the difference \(\displaystyle v-w\) by subtracting component-wise

\(\displaystyle v-w=(1,3)-(5,2)=(1-5,3-2)=(-4,1)\)

First, multiply out the vectors by their corresponding coefficients:

\(\displaystyle 3\mathbf{u}=3(4\mathbf{i}+8\mathbf{j}+12\mathbf{k})=12\mathbf{i}+24\mathbf{j}+36\mathbf{k}\)

\(\displaystyle 2\mathbf{v}=2(3\mathbf{i}+6\mathbf{j}+9\mathbf{k})=6\mathbf{i}+12\mathbf{j}+18\mathbf{k}\)

Now we simply subtract the two vectors:

\(\displaystyle 3\mathbf{u}-2\mathbf{v}=(12-6)\mathbf{i}+(24-12)\mathbf{j}+(36-18)\mathbf{k}=6\mathbf{i}+12\mathbf{j}+18\mathbf{k}\)

Set up an expression to subtract the vectors.

\(\displaystyle \left \langle -3,6,-9\right \rangle-\left \langle -1,2,-9\right \rangle\)

Be sure to make note of the sign change when subtracting the vectors.

\(\displaystyle \left \langle -3-(-1),6-2,(-9)-(-9)\right \rangle\)

Simplify the vector.

The answer is:  \(\displaystyle \left \langle -2,4,0\right \rangle\)

To find the difference, \(\displaystyle u-v\), we distribute the minus sign and turn it into vector addition.
\(\displaystyle u-v=< -3,-7>-< 6,4>=< -3,-7>+< -6,-4>\)
The sum of two vectors is defined as

The sum of two vectors 

\(\displaystyle a=< a_1, a_2>\)

\(\displaystyle b=< b_1, b_2>\)

is defined as 

\(\displaystyle a+b=< a_1+b_1, a_2+b_2>\)

As such, the difference is

\(\displaystyle < -9,-11>\)

The formula for subtracting vectors is \(\displaystyle a-b=\left \langle a_1-b_!,a_2-b_2,a_3-b_3\right \rangle\). Plugging in the values we were given, we get \(\displaystyle \left \langle (1-2),(3--4),(2-1)\right \rangle=\left \langle -1,7,1\right \rangle\)

To subtract vectors, we simply subtract their respective components:

\(\displaystyle \left \langle z-\ln(z), 0, y^2z-2\right \rangle\)

Note that our result is a vector still.

The formula for the difference of vectors is 

\(\displaystyle \left \langle x_1,x_2,x_3\right \rangle-\left \langle y_1,y_2,y_3\right \rangle=\left \langle x_1-y_1,x_2-y_2,x_3-y_3\right \rangle\)

Using the vectors given, we get 

\(\displaystyle \left \langle 2-3,0-3,6-1\right \rangle=\left \langle -1,-3,5\right \rangle\).

The formula for subtracting vectors a and b is 

\(\displaystyle a-b=\left \langle a_1-b_1,a_2-b_2,a_3-b_3\right \rangle\).

Plugging in the values we were given, we get 

\(\displaystyle \left \langle (2-3),(5-0),(19-1)\right \rangle=\left \langle -1,5,18\right \rangle\)

The formula for the difference of vectors \(\displaystyle a=\left \langle x_1,y_1,z_1\right \rangle\) and \(\displaystyle b=\left \langle x_2,y_2,z_2\right \rangle\) is \(\displaystyle a-b=\left \langle (x_1-x_2),(y_1-y_2),(z_1-z_2) \right \rangle\). Using the vectors from the problem statement, we get \(\displaystyle \left \langle (2-5),(12-5),(13-1) \right \rangle=\left \langle -3,7,12\right \rangle\)