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3.1 Vectors in 2-Space, 3-Space, and n-Space: Class Notes

Definition

If a vector $𝐯$ in 2-space or 3-space is positioned with its initial point at the origin of a rectangular coordinate system, then the coordinates of its terminal point are components of $𝐯$ relative to the coordinate system. The numbers in an $n$-tuple can be considered coordinates of a point or components of a vector.

#5

a. Find the terminal point of the vector that is equivalent to $𝐮=(1,2)$ and whose initial point is $A(1,1)$.

b. Find the initial point of the vector that is equivalent to $𝐮=(1,1,3)$ and whose terminal point is $B(-1,-1,2)$.

Solution

a. Equivalent vectors have the same length and direction. Start at $A(1,1)$ and move by the components of $\overrightarrow{u}=(1,2)$:

$$(1+1,1+2)=(2,3).$$

Coordinate axes. The vector $\overrightarrow{u}=(1,2)$ is drawn from the origin. An equivalent copy starts at $A(1,1)$ and ends at $(2,3)$; the two arrows are parallel and the same length.

Check: the copy of $\overrightarrow{u}$ has components (terminal minus initial) $(2-1,3-1)=(1,2)$.

b. (Formal approach) Let $D=(a,b,c)$ be the initial point of $\overrightarrow{u}$. Terminal minus initial gives the components:

$$\overrightarrow{u}=(-1-a,-1-b,2-c)=(1,1,3),$$

so $-1-a=1$, $-1-b=1$, and $2-c=3$. Then $a=-2$, $b=-2$, $c=-1$.

#7

Find the initial point $P$ of a nonzero vector $𝐮=\overrightarrow{PQ}$ with terminal point $Q(3,0,-5)$ and such that

a. $𝐮$ has the same direction as $𝐯=(4,-2,-1)$.

b. $𝐮$ is oppositely directed to $𝐯=(4,-2,-1)$.

Solution

a. Let $P(a,b,c)$. Then $\overrightarrow{u}=(3-a,0-b,-5-c)$. Take $\overrightarrow{u}=\overrightarrow{v}$:

$$3-a=4,0-b=-2,-5-c=-1.$$

Tip: Any positive multiple of $\overrightarrow{v}$ has the same direction, so other answers are possible; choosing $\overrightarrow{u}=\overrightarrow{v}$ gives the simplest one.

b. $\overrightarrow{u}$ oppositely directed to $\overrightarrow{v}$ means $\overrightarrow{u}=-\overrightarrow{v}=(-4,2,1)$. Then, the same as part (a):

$$3-a=-4,0-b=2,-5-c=1.$$

Definition

#12

Let $𝐮=(1,2,-3,5,0)$, $𝐯=(0,4,-1,1,2)$, and $𝐰=(7,1,-4,-2,3)$. Find the components of

a. $𝐯+𝐰$

b. $3(2𝐮-𝐯)$

c. $(3𝐮-𝐯)-(2𝐮+4𝐰)$

d. ${\displaystyle \frac{1}{2}(𝐰-5𝐯+2𝐮)+𝐯}$

Solution

a. Add matching components of $\overrightarrow{v}$ and $\overrightarrow{w}$:

$$\overrightarrow{v}+\overrightarrow{w}=(0+7,4+1,-1+(-4),1+(-2),2+3)$$

b.

$$3(2\overrightarrow{u}-\overrightarrow{v})=3(2-0,4-4,-6+1,10-1,0-2)=3(2,0,-5,9,-2)$$

c.

$$\begin{array}{cc}\hfill {\displaystyle (3\overrightarrow{u}-\overrightarrow{v})-(2\overrightarrow{u}+4\overrightarrow{w})}& {\displaystyle =(3-0,6-4,-9+1,15-1,0-2)}\hfill \\ \hfill {\displaystyle }& {\displaystyle \phantom{=}-(2+28,4+4,-6-16,10-8,0+12)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(3,2,-8,14,-2)-(30,8,-22,2,12)}\hfill \end{array}$$

Geometrical vector addition

Left: $\overrightarrow{v}$ is placed with its initial point at the tip of $\overrightarrow{u}$; $\overrightarrow{u}+\overrightarrow{v}$ goes from the tail of $\overrightarrow{u}$ to the tip of $\overrightarrow{v}$. Right: $\overrightarrow{u}$ and $\overrightarrow{v}$ share an initial point and form two sides of a parallelogram (other sides dashed); $\overrightarrow{u}+\overrightarrow{v}$ is the diagonal from the shared initial point.

Scalar multiplication

For $k>1$, $k\overrightarrow{u}$ points the same direction as $\overrightarrow{u}$ and is $k$ times as long.

Subtraction

$\overrightarrow{u}$ and $\overrightarrow{v}$ share an initial point. $\overrightarrow{u}-\overrightarrow{v}$ goes from the tip of $\overrightarrow{v}$ to the tip of $\overrightarrow{u}$, since $\overrightarrow{v}+(\overrightarrow{u}-\overrightarrow{v})=\overrightarrow{u}$.

The zero vector, $\mathrm{𝟎}=(0,0,\dots ,0)$, has length zero.

Written by hand, the zero vector is $\overrightarrow{0}$.

Theorem 3.1.1

Proof of (f)

This proof uses the components of the vectors. Let $\overrightarrow{u}=(u_1,u_2,\dots ,u_n)$ and $k,m\in R$.

$$\begin{array}{cc}\hfill {\displaystyle (k+m)\overrightarrow{u}}& {\displaystyle =((k+m)u_1,(k+m)u_2,\dots ,(k+m)u_n)}\hfill \\ \hfill {\displaystyle }& {\displaystyle \phantom{=}\text{by definition of scalar multiplication}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(k{u}_1+m{u}_1,k{u}_2+m{u}_2,\dots ,k{u}_n+m{u}_n)}\hfill \\ \hfill {\displaystyle }& {\displaystyle \phantom{=}\text{distributive property for real numbers}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(k{u}_1,k{u}_2,\dots ,k{u}_n)+(m{u}_1,m{u}_2,\dots ,m{u}_n)}\hfill \\ \hfill {\displaystyle }& {\displaystyle \phantom{=}\text{by definition of vector addition}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =k\overrightarrow{u}+m\overrightarrow{u}}\hfill \\ \hfill {\displaystyle }& {\displaystyle \phantom{=}\text{by definition of scalar multiplication}}\hfill \end{array}$$

Tip: When we use properties of real numbers, we need to use them on real numbers: here the distributive property is applied to the components $(k+m)u_i$, which are real numbers.

Theorem 3.1.2

Definition

#17

Let $𝐮=(1,-1,3,5)$ and $𝐯=(2,1,0,-3)$. Find scalars $a$ and $b$ so that $a𝐮+b𝐯=(1,-4,9,18)$.

Solution

Call the right side $\overrightarrow{c}$ and write the vectors as columns:

$$a\overrightarrow{u}+b\overrightarrow{v}=\overrightarrow{c}\Rightarrow \left[\begin{array}{c}1\\ -1\\ 3\\ 5\end{array}\right]a+\left[\begin{array}{c}2\\ 1\\ 0\\ -3\end{array}\right]b=\left[\begin{array}{c}1\\ -4\\ 9\\ 18\end{array}\right]$$

$$\Rightarrow \left[\begin{array}{c}a+2b\\ -a+b\\ 3a\\ 5a-3b\end{array}\right]=\left[\begin{array}{c}1\\ -4\\ 9\\ 18\end{array}\right]\Rightarrow \begin{array}{cc}\hfill {\displaystyle a+2b}& {\displaystyle =1}\hfill \\ \hfill {\displaystyle -a+b}& {\displaystyle =-4}\hfill \\ \hfill {\displaystyle 3a}& {\displaystyle =9}\hfill \\ \hfill {\displaystyle 5a-3b}& {\displaystyle =18}\hfill \end{array}$$

This system has augmented matrix with columns $\overrightarrow{u}$, $\overrightarrow{v}$, and $\overrightarrow{c}$. Row reduce:

$$\left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -4}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 9}\hfill \\ \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 18}\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 3}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Check:

$$3(1,-1,3,5)-(2,1,0,-3)=(3-2,-3-1,9-0,15+3)=(1,-4,9,18).$$

Tip: This is overkill, but it lets us practice a matrix approach.