# 2.2 Vectors in three dimensions  (Page 6/14)

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Let $\text{v}=⟨-1,-1,1⟩$ and $\text{w}=⟨2,0,1⟩.$ Find a unit vector in the direction of $5\text{v}+3\text{w}.$

$⟨\frac{1}{3\sqrt{10}},-\frac{5}{3\sqrt{10}},\frac{8}{3\sqrt{10}}⟩$

## Throwing a forward pass

A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of $30\text{°}$ (see the following figure). Write the initial velocity vector of the ball, $\text{v},$ in component form.

The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector $\text{w}$ extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of $30\text{°}$ (see the following figure). This vector would have the same direction as $\text{v},$ but it may not have the right magnitude.

The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is

$\text{Dist from QB to receiver}=\sqrt{{15}^{2}+{20}^{2}}=\sqrt{225+400}=\sqrt{625}=25\phantom{\rule{0.2em}{0ex}}\text{yd}.$

We have $\frac{25}{‖\text{w}‖}=\text{cos}\phantom{\rule{0.2em}{0ex}}30\text{°}.$ Then the magnitude of $\text{w}$ is given by

$‖\text{w}‖=\frac{25}{\text{cos}\phantom{\rule{0.2em}{0ex}}30\text{°}}=\frac{25·2}{\sqrt{3}}=\frac{50}{\sqrt{3}}\phantom{\rule{0.2em}{0ex}}\text{yd}$

and the vertical distance from the receiver to the terminal point of $\text{w}$ is

$\text{Vert dist from receiver to terminal point of}\phantom{\rule{0.2em}{0ex}}\text{w}=‖\text{w}‖\text{sin}\phantom{\rule{0.2em}{0ex}}30\text{°}=\frac{50}{\sqrt{3}}·\frac{1}{2}=\frac{25}{\sqrt{3}}\phantom{\rule{0.2em}{0ex}}\text{yd}.$

Then $\text{w}=⟨20,15,\frac{25}{\sqrt{3}}⟩,$ and has the same direction as $\text{v}.$

Recall, though, that we calculated the magnitude of $\text{w}$ to be $‖\text{w}‖=\frac{50}{\sqrt{3}},$ and $\text{v}$ has magnitude $60$ mph. So, we need to multiply vector $\text{w}$ by an appropriate constant, $k.$ We want to find a value of $k$ so that $‖k\text{w}‖=60$ mph. We have

$‖k\text{w}‖=k‖\text{w}‖=k\frac{50}{\sqrt{3}}\phantom{\rule{0.2em}{0ex}}\text{mph,}$

so we want

$\begin{array}{}\\ \\ \hfill k\frac{50}{\sqrt{3}}& =\hfill & 60\hfill \\ \hfill k& =\hfill & \frac{60\sqrt{3}}{50}\hfill \\ \hfill k& =\hfill & \frac{6\sqrt{3}}{5}.\hfill \end{array}$

Then

$\text{v}=k\text{w}=k⟨20,15,\frac{25}{\sqrt{3}}⟩=\frac{6\sqrt{3}}{5}⟨20,15,\frac{25}{\sqrt{3}}⟩=⟨24\sqrt{3},18\sqrt{3},30⟩.$

Let’s double-check that $‖\text{v}‖=60.$ We have

$‖\text{v}‖=\sqrt{{\left(24\sqrt{3}\right)}^{2}+{\left(18\sqrt{3}\right)}^{2}+{\left(30\right)}^{2}}=\sqrt{1728+972+900}=\sqrt{3600}=60\phantom{\rule{0.2em}{0ex}}\text{mph}.$

So, we have found the correct components for $\text{v}.$

Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of $40$ mph and an angle of $45\text{°}.$ Write the initial velocity vector of the ball, $\text{v},$ in component form.

$\text{v}=⟨16\sqrt{2},12\sqrt{2},20\sqrt{2}⟩$

## Key concepts

• The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples $\left(x,y,z\right)$ are used to describe the location of a point in space.
• The distance $d$ between points $\left({x}_{1},{y}_{1},{z}_{1}\right)$ and $\left({x}_{2},{y}_{2},{z}_{2}\right)$ is given by the formula
$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}+{\left({z}_{2}-{z}_{1}\right)}^{2}}.$
• In three dimensions, the equations $x=a,y=b,\text{and}\phantom{\rule{0.2em}{0ex}}z=c$ describe planes that are parallel to the coordinate planes.
• The standard equation of a sphere with center $\left(a,b,c\right)$ and radius $r$ is
${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}+{\left(z-c\right)}^{2}={r}^{2}.$
• In three dimensions, as in two, vectors are commonly expressed in component form, $\text{v}=⟨x,y,z⟩,$ or in terms of the standard unit vectors, $x\text{i}+y\text{j}+z\text{k}.$
• Properties of vectors in space are a natural extension of the properties for vectors in a plane. 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: $\frac{\text{v}}{‖\text{v}‖}=\frac{1}{‖\text{v}‖}⟨{x}_{1},{y}_{1},{z}_{1}⟩=⟨\frac{{x}_{1}}{‖\text{v}‖},\frac{{y}_{1}}{‖\text{v}‖},\frac{{z}_{1}}{‖\text{v}‖}⟩,$ $\text{v}\ne 0$

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