Vectors in coordinates

The unit vectors of the axes are $\displaystyle \vec{i}$, $\displaystyle \vec{j}$, $\displaystyle \vec{k}$: each is 1 long, and their directions are those of the $\displaystyle x$, $\displaystyle y$, $\displaystyle z$ axes.

The three are not in one plane, so every vector is a unique linear combination of them (the theorem of lesson 10):

This is $\displaystyle \vec{u}$ in coordinates – an ordered triple of numbers. The vector from $\displaystyle O$ to the point $\displaystyle P(x,y,z)$ is $\displaystyle (x,y,z)$: in the drawing $\displaystyle {\vec{u}}={\overrightarrow{OP}}={(2,3,2)}$.

$\displaystyle {\overrightarrow{AB}}={\overrightarrow{OB}-\overrightarrow{OA}}$, so – the end minus the start:

And operations on vectors work coordinate by coordinate:

• Equality: two vectors are equal if and only if all their coordinates are equal.
• Adding: $\displaystyle {(x_1,y_1,z_1)+(x_2,y_2,z_2)}={(x_1+x_2,\ y_1+y_2,\ z_1+z_2)}$.
• Multiplying by a number: $\displaystyle {t(x,y,z)}={(tx,ty,tz)}$.
• Collinearity: $\displaystyle {\vec{v}}={t\vec{u}}$ – each coordinate of $\displaystyle \vec{v}$ is the same number $\displaystyle t$ times that of $\displaystyle \vec{u}$.
• The length – a box’s diagonal, as in the last lesson:

The points $\displaystyle A(3,-4,1)$, $\displaystyle B(4,1,8)$, $\displaystyle C(2,0,7)$ are given.

• $\displaystyle {\overrightarrow{AB}}={(1,5,7)}$, $\displaystyle {\overrightarrow{AC}}={(-1,4,6)}$.
• They are not collinear: $\displaystyle {-1}={t\cdot1}$ gives $\displaystyle {t}={-1}$, but $\displaystyle 4\neq-1\cdot5$. So $\displaystyle A$, $\displaystyle B$, $\displaystyle C$ are not on one line – they make a triangle.
• $\displaystyle {|\overrightarrow{AB}|}={\sqrt{1+25+49}}={\sqrt{75}}={5\sqrt{3}}$.
• Where the medians meet, $\displaystyle M$: $\displaystyle {\overrightarrow{AM}}={\frac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)}={\frac{1}{3}(0,9,13)}$, so $\displaystyle {M}={\left(3,\ -4+3,\ 1+\frac{13}{3}\right)}={\left(3,-1,\frac{16}{3}\right)}$.
• The midpoint of a segment is the average of the coordinates: the midpoint of $\displaystyle AB$ is $\displaystyle {\left(\frac{3+4}{2},\frac{-4+1}{2},\frac{1+8}{2}\right)}={(3.5,-1.5,4.5)}$.

Is $\displaystyle {\vec{w}}={(3,6,8)}$ a linear combination of $\displaystyle {\vec{u}}={(1,5,7)}$ and $\displaystyle {\vec{v}}={(-1,4,6)}$?

$\displaystyle {s\vec{u}+t\vec{v}}={\vec{w}}$ is an equality of three coordinates – three equations in two unknowns:

From the first two: $\displaystyle {s}={2}$, $\displaystyle {t}={-1}$. They satisfy the third as well: $\displaystyle {14-6}={8}$. So $\displaystyle {\vec{w}}={2\vec{u}-\vec{v}}$, and $\displaystyle \vec{w}$ is in the plane of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$. Had the third equation failed, the system would have no solution, and $\displaystyle \vec{w}$ would be no combination of them.

The points $\displaystyle A(1,2,-1)$ and $\displaystyle B(4,-2,11)$ are given.

Write $\displaystyle \overrightarrow{AB}$ in coordinates.

The vectors $\displaystyle {\vec{u}}={(2,-1,3)}$ and $\displaystyle {\vec{v}}={(1,4,-2)}$ are given.

Find $\displaystyle 2\vec{u}-3\vec{v}$.

$\displaystyle ABCD$ is a parallelogram: $\displaystyle A(1,0,2)$, $\displaystyle B(4,1,3)$, $\displaystyle C(5,3,-1)$.
Find the coordinates of $\displaystyle D$. (Hint: $\displaystyle {\overrightarrow{AD}}={\overrightarrow{BC}}$.)

Are the points $\displaystyle A(1,1,1)$, $\displaystyle B(3,2,-1)$, $\displaystyle C(7,4,-5)$ on one line? Explain with vectors.

$\displaystyle {\vec{u}}={(1,2,1)}$, $\displaystyle {\vec{v}}={(2,3,-1)}$, $\displaystyle {\vec{w}}={(4,7,1)}$.
Find $\displaystyle s$ and $\displaystyle t$ so that $\displaystyle {\vec{w}}={s\vec{u}+t\vec{v}}$: write the three equations and solve.

$\displaystyle {\vec{u}}={(1,0,1)}$, $\displaystyle {\vec{v}}={(0,1,1)}$, $\displaystyle {\vec{w}}={(1,1,1)}$.
Here is the system of $\displaystyle {\vec{w}}={s\vec{u}+t\vec{v}}$. Solve it: is $\displaystyle \vec{w}$ a linear combination of $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$?