# 2.2 Vectors in three dimensions  (Page 5/14)

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Three-dimensional vectors can also be represented in component form. The notation $\text{v}=⟨x,y,z⟩$ is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, $\left(0,0,0\right),$ and terminal point $\left(x,y,z\right).$ The zero vector is $0=⟨0,0,0⟩.$ So, for example, the three dimensional vector $\text{v}=⟨2,4,1⟩$ is represented by a directed line segment from point $\left(0,0,0\right)$ to point $\left(2,4,1\right)$ ( [link] ). Vector v = ⟨ 2 , 4 , 1 ⟩ is represented by a directed line segment from point ( 0 , 0 , 0 ) to point ( 2 , 4 , 1 ) .

Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If $\text{v}=⟨{x}_{1},{y}_{1},{z}_{1}⟩$ and $\text{w}=⟨{x}_{2},{y}_{2},{z}_{2}⟩$ are vectors, and $k$ is a scalar, then

$\text{v}+\text{w}=⟨{x}_{1}+{x}_{2},{y}_{1}+{y}_{2},{z}_{1}+{z}_{2}⟩\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}k\text{v}=⟨k{x}_{1},k{y}_{1},k{z}_{1}⟩.$

If $k=-1,$ then $k\text{v}=\left(-1\right)\text{v}$ is written as $\text{−}\text{v},$ and vector subtraction is defined by $\text{v}-w=v+\left(\text{−}\text{w}\right)=v+\left(-1\right)\text{w}.$

The standard unit vectors extend easily into three dimensions as well— $\text{i}=⟨1,0,0⟩,$ $\text{j}=⟨0,1,0⟩,$ and $\text{k}=⟨0,0,1⟩$ —and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in ${ℝ}^{3}$ in the following ways:

$\text{v}=⟨x,y,z⟩=x\text{i}+y\text{j}+z\text{k}.$

## Vector representations

Let $\stackrel{\to }{PQ}$ be the vector with initial point $P=\left(3,12,6\right)$ and terminal point $Q=\left(-4,-3,2\right)$ as shown in [link] . Express $\stackrel{\to }{PQ}$ in both component form and using standard unit vectors. The vector with initial point P = ( 3 , 12 , 6 ) and terminal point Q = ( −4 , −3 , 2 ) .

In component form,

$\begin{array}{cc}\hfill \stackrel{\to }{PQ}& =⟨{x}_{2}-{x}_{1},{y}_{2}-{y}_{1},{z}_{2}-{z}_{1}⟩\hfill \\ & =⟨-4-3,-3-12,2-6⟩=⟨-7,-15,-4⟩.\hfill \end{array}$

In standard unit form,

$\stackrel{\to }{PQ}=-7\text{i}-15\text{j}-4\text{k}.$

Let $S=\left(3,8,2\right)$ and $T=\left(2,-1,3\right).$ Express $\stackrel{\to }{ST}$ in component form and in standard unit form.

$\stackrel{\to }{ST}=⟨-1,-9,1⟩=\text{−}\text{i}-9\text{j}+\text{k}$

As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space ( [link] ).

We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.

## Rule: properties of vectors in space

Let $\text{v}=⟨{x}_{1},{y}_{1},{z}_{1}⟩$ and $\text{w}=⟨{x}_{2},{y}_{2},{z}_{2}⟩$ be vectors, and let $k$ be a scalar.

Scalar multiplication: $k\text{v}=⟨k{x}_{1},k{y}_{1},k{z}_{1}⟩$

Vector addition: $\text{v}+\text{w}=⟨{x}_{1},{y}_{1},{z}_{1}⟩+⟨{x}_{2},{y}_{2},{z}_{2}⟩=⟨{x}_{1}+{x}_{2},{y}_{1}+{y}_{2},{z}_{1}+{z}_{2}⟩$

Vector subtraction: $\text{v}-\text{w}=⟨{x}_{1},{y}_{1},{z}_{1}⟩-⟨{x}_{2},{y}_{2},{z}_{2}⟩=⟨{x}_{1}-{x}_{2},{y}_{1}-{y}_{2},{z}_{1}-{z}_{2}⟩$

Vector magnitude: $‖\text{v}‖=\sqrt{{x}_{1}{}^{2}+{y}_{1}{}^{2}+{z}_{1}{}^{2}}$

Unit vector in the direction of v: $\frac{1}{‖\text{v}‖}\text{v}=\frac{1}{‖\text{v}‖}⟨{x}_{1},{y}_{1},{z}_{1}⟩=⟨\frac{{x}_{1}}{‖\text{v}‖},\frac{{y}_{1}}{‖\text{v}‖},\frac{{z}_{1}}{‖\text{v}‖}⟩,$ if $\text{v}\ne 0$

We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

## Vector operations in three dimensions

Let $\text{v}=⟨-2,9,5⟩$ and $\text{w}=⟨1,-1,0⟩$ ( [link] ). Find the following vectors.

1. $3\text{v}-2\text{w}$
2. $5‖\text{w}‖$
3. $‖5\text{w}‖$
4. A unit vector in the direction of $\text{v}$ The vectors v = ⟨ −2 , 9 , 5 ⟩ and w = ⟨ 1 , −1 , 0 ⟩ .
1. First, use scalar multiplication of each vector, then subtract:
$\begin{array}{cc}\hfill 3\text{v}-2\text{w}& =3⟨-2,9,5⟩-2⟨1,-1,0⟩\hfill \\ & =⟨-6,27,15⟩-⟨2,-2,0⟩\hfill \\ & =⟨-6-2,27-\left(-2\right),15-0⟩\hfill \\ & =⟨-8,29,15⟩.\hfill \end{array}$
2. Write the equation for the magnitude of the vector, then use scalar multiplication:
$5‖\text{w}‖=5\sqrt{{1}^{2}+{\left(-1\right)}^{2}+{0}^{2}}=5\sqrt{2}.$
3. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:
$‖5\text{w}‖=‖⟨5,-5,0⟩‖=\sqrt{{5}^{2}+{\left(-5\right)}^{2}+{0}^{2}}=\sqrt{50}=5\sqrt{2}.$
4. Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:
$\begin{array}{cc}\hfill \frac{\text{v}}{‖\text{v}‖}& =\frac{1}{‖\text{v}‖}⟨-2,9,5⟩\hfill \\ & =\frac{1}{\sqrt{{\left(-2\right)}^{2}+{9}^{2}+{5}^{2}}}⟨-2,9,5⟩\hfill \\ & =\frac{1}{\sqrt{110}}⟨-2,9,5⟩\hfill \\ & =⟨\frac{-2}{\sqrt{110}},\frac{9}{\sqrt{110}},\frac{5}{\sqrt{110}}⟩.\hfill \end{array}$

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