# Introduction to Vectors

This section covers:

# Introduction to Vectors

A vector (also called a direction vector) is just is something that has both magnitude (length, or size) and direction. It’s different than a regular number, since it really has two components to it. We see vectors represented by arrows, so we can remember that we need to get a length of a vector (the magnitude), as well as the direction (which way it’s pointing).

We use vectors in mathematics, engineering, and physics, since many times we need to know both the size of something and which way it’s going. For example, with an airplane, we can use a vector to measure the speed of the plane (the “size”) and the direction it’s flying.

Geometric Vectors are directed line segments in the $$xy$$-plane, and, as an example, the vector from a point $$A$$ (initial point) to a point $$B$$ (terminal point) can be represented by $$\overrightarrow{{AB}}$$ .

For example, if $$A$$ is $$(2,7)$$ and $$B$$ is $$(-3,8)$$, the vector  is second point minus first point, or $$\displaystyle \left\langle {{{x}_{2}}} \right.-{{x}_{1}},\left. {{{y}_{2}}-{{y}_{1}}} \right\rangle$$, or $$\left\langle {-3-2} \right.,\left. {8-7} \right\rangle =\left\langle {-5,\left. 1 \right\rangle } \right.$$. The “$$x$$” part of the vector (–5) is called the $$x$$-component, and the ”$$y$$” part (1) is called the $$y$$-component. This $$\left\langle {x,} \right.\left. y \right\rangle$$ form is called component form. We usually call vectors with single letters, like $$\overrightarrow{\text{u}}$$, $$\overrightarrow{\text{v}}$$ or $$\overrightarrow{\text{w}}$$, or just u, v, w.

Note also that vectors can also be written in the form $$\text{ai}+\text{bj}$$ (called the linear combination of the unit vectors $$\text{i}$$ and $$\text{j}$$), so this vector can also be written as $$-5\text{i}+1\text{j}$$, or $$-5\text{i}+\text{j}$$.

The magnitude of the vector, written $$\left\| {AB} \right\|$$ is the distance between the two points (like the hypotenuse of a right triangle), or $$\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}$$, or with the new vector $$\left\langle {x,} \right.\left. y \right\rangle$$, it’s just $$\sqrt{{{{x}^{2}}+{{y}^{2}}}}$$. For our points $$A$$ and $$B$$ above, $$\left\| {AB} \right\|=\sqrt{{{{{\left( {-5} \right)}}^{2}}+{{1}^{2}}}}=\sqrt{{26}}$$.

Now looking at this vector visually, do you see how we can use the slope of the line of the vector (from the initial point to the terminal point) to get the direction of the vector? Pretty cool! We can just use $$\displaystyle {{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)$$ (second part of vector over first part of vector) to get the angle measurement of the vector’s direction. Remembering from the Polar Coordinates, Equations and Graphs section though, we have to be careful which quadrant the vector terminates in (“pretending” that the vector’s initial point is at the origin) to know how many degrees we should add to that tangent value when we use a calculator:

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Here is all this visually. Note that we had to add 180° to the angle measurement we got from the calculator (–11.3°) since the vector would terminate in the 2nd quadrant if we were to start at $$(0,0)$$. We get 168.7°, which is the angle measurement from the positive $$\boldsymbol {x}$$-axis going counterclockwise.

168.7° from the positive $$x$$-axis can also be described as 11.3° North of West (11.3° N of W, or W11.3°W), since the closest axis to the angle is the negative $$x$$-axis (west) and we are going a little north of that:

(We saw a similar concept of this when we were working with bearings here in the Law of Sines and Cosines, and Areas of Triangles section).

Note that a vector that has a magnitude of 0 (and thus no direction) is called a zero vector. Thus, hypothetically, the vector $$\overrightarrow{{AA}}$$ would be a zero vector.

A unit vector is a vector with magnitude 1; in some applications, it’s easier to work with unit vectors. To find the unit vector that is associated with a vector (has same direction, but magnitude of 1), use the following formula: $$\displaystyle \text{u}=\frac{\text{v}}{{\left\| \text{v} \right\|}}$$ (just divide each component of a vector by its magnitude to get its unit vector). We’ll see some problems below.

# Vector Operations

## Adding and Subtracting Vectors

There are a couple of ways to add and subtract vectors. When we add vectors, geometrically, we just put the beginning point (initial point) of the second vector at the end point (terminal point) of the first vector, and see where we end up (new vector starts at beginning of one and ends at end of the other). If the vectors aren’t this way to begin with, we can move the second vector (as long as it has the same magnitude and direction, so it’s like a slide) to be this way. This is called the “head-to-tail” method.

You can think of adding vectors as connecting the diagonal of the parallelogram (a four-sided figure with two pairs of parallel sides) that contains the two vectors.

Do you see how when we add vectors geometrically, to get the sum, we can just add the $$x$$ components of the vector, and the $$y$$ components of the vectors?

When we subtract two vectors, we just take the vector that’s being subtracting, reverse the direction and add it to the first vector. This is because the negative of a vector is that vector with the same magnitude, but has an opposite direction (thus adding a vector and its negative results in a zero vector).

Note that to make a vector negative, you can just negate each of its components ($$x$$ component and $$y$$ component) (see graph below).

## Multiplying Vectors by a Number (Scalar)

To multiply a vector by a number, or scalar, you simply stretch (or shrink if the absolute value of that number is less than 1), or you can simply multiply the $$x$$ component and $$y$$ component by that number. Notice also that the magnitude is multiplied by that scalar. Multiplying by a negative number also changes the direction of that vector.

Can you see how two vectors that are parallel are always a multiple of each other (with a multiple of 1 if the vectors are the same size)?

Here’s what subtracting vectors and also multiplying vectors by a scalar looks like:

Let’s put all this together to perform the following vector operations, given the vectors shown:

 EXAMPLE VECTORS Vector Operations Vector Operations $$\begin{array}{l}\text{u}:\,\,\left\langle {4,1} \right\rangle =4\text{i}+\text{j}\\\text{v}:\,\,\left\langle {-2,2} \right\rangle =-2\text{i}+2\text{j}\\\text{w}:\,\,\left\langle {0,6} \right\rangle =6\text{j}\end{array}$$ $$\text{u}-2\text{v}+\text{w}$$ $$\left\langle {4,1} \right\rangle -2\left\langle {-2,2} \right\rangle +\left\langle {0,6} \right\rangle =\left\langle {8,3} \right\rangle$$ or $$\left( {4\text{i}+\text{j}} \right)-2\left( {-2\text{i}+2\text{j}} \right)+6\text{j}=8\text{i}+3\text{j}$$ $$\text{u}+3\text{v}$$ $$\left\langle {4,1} \right\rangle +3\left\langle {-2,2} \right\rangle =\left\langle {-2,7} \right\rangle$$ or $$\left( {4\text{i}+\text{j}} \right)+3\left( {-2\text{i}+2\text{j}} \right)=-2\text{i}+7\text{j}$$

You may also see problems like this, where you have to tell whether the statement is true or false. Note that you want to look at where you end up in relation to where you started to see the resulting vector. If you end up exactly where you started from, the resulting vector is 0.

 Vectors Graphically True or False (a)    A + B = C?    True   (b)   A + B – C = 0?    True   (c)    B =  2G?    False  (B = –.5G)   (d)   A + B + D + G = –A = –E?    True (We end up “E” down from the initial point of A; A and E are the same)   (e)    C + F – G = D?    False  (Even though we end up at the terminal point of D, D doesn’t start from initial point of C)   (f)    B – C – E = –2E = –2A?   True

Here are a couple more examples of vector problems. Notice in the second set of problems when we are given a magnitude and direction of a vector, and have to find that vector, we use the following equation, like we did when we here in the Polar Coordinates, Equations and Graphs section, where $$\left\| {\text{v}} \right\|$$ or the magnitude of a vector is like the “$$r$$” (radius) we saw for polar numbers: $$\text{v}=\left\| {\text{v}} \right\|\left( {\cos \alpha \text{i}+\sin \alpha \text{j}} \right)=\left( {\left\| {\text{v}} \right\|\cos \alpha } \right)\text{i}+\left( {\left\| {\text{v}} \right\|\sin \alpha } \right)\text{j}$$. (Trigonometry always seems to come back and haunt us!) We’ll leave our answers in $$\text{a}\text{i}+\text{b}\text{j}$$ form.

 Vector Problems Solutions For initial point P and terminal point Q, write vector v in from $$\text{a}\text{i}+\text{b}\text{j}$$ (its position vector):   (a) $$P=\left( {4,-2} \right)\,\,\,\,Q=\left( {0,4} \right)$$ (b) $$P=\left( {0,20} \right)\,\,\,\,\,Q=\left( {-1,10} \right)$$   Find the direction for (b). Subtract initial point from terminal point: (a) $$\displaystyle \text{v}=\overrightarrow{{PQ}}=\left\langle {0-4,} \right.\left. {4-\left( {-2} \right)} \right\rangle =\left\langle {-4,6} \right\rangle =-4\text{i}+6\text{j}$$ (b) $$\displaystyle \text{v}=\overrightarrow{{PQ}}=\left\langle {-1-0,} \right.\left. {10-20} \right\rangle =\left\langle {-1,-10} \right\rangle =-\text{i}-10\text{j}$$   For the direction of $$\displaystyle -\text{i}-10\text{j}$$, we have to use $$\displaystyle {{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)$$; we get 84.3° from the calculator, but this vector ends up in the third quadrant. We then have to add 180° to get an angle from the positive $$x$$-axis of 264.3°. Perform the “reverse” math to get to get vectors back:   Find a vector v in the form $$\text{a}\text{i}+\text{b}\text{j}$$ given its magnitude and the angle it makes with the positive $$x$$-axis:   (a) $$\left\| {\text{v}} \right\|=4,\,\,\,\,\,\alpha =135{}^\circ$$ (b) $$\displaystyle \left\| {\text{v}} \right\|=2,\,\,\,\,\,\alpha =\frac{{11\pi }}{6}$$ We’ll need to use $$\text{v}=\left\| {\,\text{v}} \right\|\left( {\cos \alpha \text{i}+\sin \alpha \text{j}} \right)$$ to get our $$x$$ and $$y$$ components of the vector, respectively:   (a) \begin{align}\text{v}&=\left\| {\operatorname{v}} \right\|\left( {\cos \alpha \text{i}+\sin \alpha \text{j}} \right)=4\cos \left( {135} \right)\text{i}+4\sin \left( {135} \right)\text{j}\\&=4\left( {\frac{{-\sqrt{2}}}{2}} \right)\text{i}+4\left( {\frac{{\sqrt{2}}}{2}} \right)\text{j}=-2\sqrt{2}\text{i}+2\sqrt{2}\text{j}\end{align}   (b) \begin{align}\text{v}&=\left\| {\text{v}} \right\|\left( {\cos \alpha \text{i}+\sin \alpha \text{j}} \right)=2\cos \left( {\frac{{11\pi }}{6}} \right)\text{i}+2\sin \left( {\frac{{11\pi }}{6}} \right)\text{j}\\&=2\left( {\frac{{\sqrt{3}}}{2}} \right)\text{i}+2\left( {-\frac{1}{2}} \right)\text{j}=\sqrt{3}\text{i}-\text{j}\end{align} Find the unit vector having the same direction as v:   (a) $$\text{v}=6\text{i}$$ (b) $$\text{v}=2\text{i}-\text{j}$$ We’ll need to use $$\displaystyle \text{u}=\frac{\text{v}}{{\left\| \text{v} \right\|}}$$ to get unit vectors:   (a) $$\displaystyle \text{u}=\frac{\text{v}}{{\left\| \text{v} \right\|}}=\frac{{6\text{i}}}{{\left\| {6\text{i}} \right\|}}=\frac{{6\text{i}}}{{\sqrt{{{{6}^{2}}+{{0}^{2}}}}}}=\frac{{6\text{i}}}{6}=\text{i}$$   (b) \displaystyle \begin{align}\text{u}&=\frac{\text{v}}{{\left\| \text{v} \right\|}}=\frac{{2\text{i}-\text{j}}}{{\left\| {2\text{i}-\text{j}} \right\|}}=\frac{{2\text{i}-\text{j}}}{{\sqrt{{{{2}^{2}}+{{{\left( {-1} \right)}}^{2}}}}}}=\frac{{2\text{i}-\text{j}}}{{\sqrt{5}}}=\frac{2}{{\sqrt{5}}}\text{i}-\frac{1}{{\sqrt{5}}}\text{j}\\&=\frac{{2\sqrt{5}}}{5}\text{i}-\frac{{\sqrt{5}}}{5}\text{j}\end{align}   Note that we rationalized the fractions in (b).

# Applications of Vectors

Vectors are extremely important in many applications of science and engineering. Since vectors include both a length and a direction, many vector applications have to do with vehicle motion and direction.

We saw above that, given a magnitude and direction, we can find the vector $$\left\langle {\left\| \text{v} \right\|\cos \theta ,\,\,\left\| \text{v} \right\|\sin \theta } \right\rangle$$, where $$\left\| \text{v} \right\|$$ is the speed. This way we can add and subtract vectors, and get a resulting speed and direction for the new vector.

Remember that a bearing (we saw here in the Law of Sines and Cosines, and Areas of Triangles section), is typically expressed a measure of the clockwise angle that starts due north or on the positive $$\boldsymbol{y}$$axis (initial side) and terminates a certain number of degrees (terminal side) from that due north starting place. (This is also written, as in the case of a bearing of 40° as “40° east of north”, or “N40°E”).

(A lot of times, the bearing includes more directions, such as 70° west of north, also written as N70°W. In this case, the angle will start due north (straight up, or on the positive $$y$$ axis) and go counterclockwise 70° (because it’s going west, or to the left, instead of east). Similarly, a bearing of 50° south of east, or E50°S, would be an angle that starts due east (on the positive $$x$$-axis) and go clockwise 50° clockwise (towards the south, or down). Also, if you see a bearing of southwest, for example, the angle would be 45° south of west, or 225° clockwise from north, and so on.)

Each time a moving object changes course, you have to draw another line to the north to map its new bearing.

When there’s a tail wind, remember that you have to add this vector to the vector that the object is trying to go on (its programmed or “steered” course), to get the actual vector of the object. Remember:

ACTUAL COURSE = PROGRAMMED COURSE + COURSE of WIND
PROGRAMMED COURSE = ACTUAL COURSE – COURSE of WIND

Here are some problems:

 Vector Application Problem Solution A plane is flying on a bearing of 25° south of west at 500 miles per hour (speed). Express the velocity of the plane (as a vector).   First, draw a picture: First draw the vector in the coordinate system to see how to get the counter-clockwise angle from the positive $$x$$–axis.   25° south of west means to go due west (negative $$x$$–axis), and then go south (down) 25°. This angle is the same (co-terminal) of the angle that is $$180°+25°=205°$$ counter-clockwise from the positive $$x$$–axis.   The speed is the same as the magnitude of the vector, so we can express the velocity vector in the form: $$\left\langle {\left\| \text{v} \right\|\cos \theta ,\left\| \text{v} \right\|\sin \theta } \right\rangle :\left\langle {500\cos 205{}^\circ ,500\sin 205{}^\circ } \right\rangle =\left\langle {-453.154,-211.309} \right\rangle$$. (Remember to put your calculator in DEGREE mode!)   Do you see how this coordinate point looks correct (3rd quadrant)? A sailboat is sailing on a bearing of 30° north of west at 6.5 miles per hour (in still water). A tail wind blowing 20 miles per hour in the direction 40° south of west alters the course of the boat.   Express the actual velocity of the sailboat as a vector. Then determine the actual speed and direction of the boat.   First, draw a picture: First draw the vectors in the coordinate system to see how to get the counter-clockwise angles from the positive $$x$$-axis.   30° north of west means to go due west (negative $$x$$-axis), and then go north (up) 30°, which has an angle of 150° (180° – 30°) from the positive $$x$$-axis. The wind, which is 40° south of west, will have an angle of 220° (180° + 40°) from the positive $$x$$-axis.   To get the vector for the  actual course of the sailboat, add the course the boat is being “steered” (programmed course) to the course of the wind. Note that we have to put the numbers in the calculator; we can’t add the velocities and angles separately: $$\begin{array}{l}\,\,\,\,\,\,\left\langle {6.5\cos 150{}^\circ ,\,\,6.5\sin 150{}^\circ } \right\rangle \,\,\,\,\,\text{Boat }\!\!’\!\!\text{ s Course}\\\underline{{+\,\,\,\left\langle {20\cos 220{}^\circ ,\,\,20\sin 220{}^\circ } \right\rangle }}\,\,\,\,\,\,\text{Wind }\!\!’\!\!\text{ s Course}\\=\,\,\,\,\,\,\,\,\left\langle {-20.950,-9.606} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Actual Course}\end{array}$$ Do you see how this coordinate point looks correct (3rd quadrant)?  (See new vector in gray).   To get the actual speed, we have to take the magnitude of this new vector: $$\sqrt{{{{{\left( {-20.95} \right)}}^{2}}+{{{\left( {-9.606} \right)}}^{2}}}}=23.047$$ miles per hour. To get the direction of the vector, use $$\displaystyle {{\tan }^{{-1}}}\left( {\frac{{-9.606}}{{-20.95}}} \right)=24.6{}^\circ +180{}^\circ =204.6{}^\circ$$. (We needed to add 180° since the vector terminates in the 3rd quadrant).   Note that if we were given a vector for the actual course of the boat and had to come up with the vector for which the boat should be “steered”, we would have to subtract the wind from the actual course.

Now let’s the problem we already did using Law of Cosines (Trig) from the Law of Sines and Cosines, and Areas of Triangles section: it is probably easier doing this problem with Trig.

Also notice that we need to use a definition of navigation bearing with respect to vectors: It is defined as the positive angle (0 to 360 degrees) measured clockwise with respect to the north (positive $$y$$-axis).

 Vector Application Problem Solution A cruise ship travels at a bearing of 40° (east of north) at 20 mph for 3 hours, and changes course to a bearing of 120° (east of north). It then travels 25 mph for 2 hours.   a) Find the distance the ship is from its original position. b) Find the ship’s new bearing from the original position.  c) On what bearing must the ship travel to return back to its original position?   First, draw a picture:   Note that we get 60 miles by multiplying rate x time to get distance, or 20 mph · 3 hours. Similarly, we get 50 miles by multiplying 25 mph by 2 hours. Since no specific directions (like west of south) are given for these bearings, we will obtain the angles by measuring the clockwise angle that starts due north or on the positive $$\boldsymbol {y}$$-axis (east of north).   The boat starts out at a 40° bearing (angle clockwise from the positive $$y$$-axis, so the angle from the positive $$x$$-axis is $$\displaystyle 90{}^\circ -40{}^\circ =50{}^\circ$$). With a change of course, we have to draw another line to the north to map its new bearing. The second bearing is 120°, so the part of this angle underneath the $$x$$-axis is 30° $$\displaystyle (120{}^\circ -90{}^\circ =30{}^\circ )$$; we can get the counterclockwise angle from the positive $$x$$–axis by subtracting 30° from 360° to get 330°. We will use this to define the vector for this second bearing.   Use vector addition to get the new vector from the starting point to the ending point. Note again that we use angle measurements by going counter-clockwise from the positive $$\boldsymbol {x}$$-axis:   $$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\left\langle {60\cos 50{}^\circ ,\,\,60\sin 50{}^\circ } \right\rangle \\\underline{{+\,\left\langle {50\cos 330{}^\circ ,\,\,50\sin 330{}^\circ } \right\rangle }}\\\approx\,\,\,\,\,\,\,\,\,\,\left\langle {81.869,\,\,20.963} \right\rangle \end{array}$$ Do you see how this coordinate point looks correct (1st quadrant) with respect to the starting point? (New vector line is dashed).   a) To get the distance between the starting point and ending point, take the magnitude of this new vector: $$\sqrt{{{{{81.869}}^{2}}+{{{20.963}}^{2}}}}\approx 84.510$$ miles.   b) To get the bearing of the ship from the original position, get the direction of the vector first: $$\displaystyle X={{\tan }^{{-1}}}\left( {\frac{{20.963}}{{81.869}}} \right)\approx 14.4{}^\circ$$. (This is also the same as 14.4° north of east.) To get the bearing from the positive $$y$$-axis going clockwise (east of north), subtract this from 90° to get 75.6°.   c) To get the bearing that the ship must travel to return back to its original position, subtact 14.4° from 270° to get 255° (east of north), which is the same as 14.4° south of west.   This is what we got when we did the problem using Law of Cosines here!

# Dot Product and Angle Between Two Vectors

The dot product of two vectors $$\text{u}=\text{ai}+\text{bj}$$ and $$\text{v}=\text{ci}+\text{dj}$$ (sort of like multiplying two vectors) is defined as $$\text{u}\bullet \text{v}=\text{ac}+\text{bd}$$; in other words, you multiply the two “$$x$$” parts of the vectors, and multiply the two “$$y$$” parts, and then add them together. The result is a scalar (single number).

Here is an example: If $$\displaystyle \text{u}=-2\text{i}+3\text{j}$$ and $$\text{v}=2\text{i}+\text{j}$$, the dot product $$\text{u}\bullet \text{v}=\left( {-2} \right)\left( 2 \right)+\left( 3 \right)\left( 1 \right)=-1$$.

We use dot products to find the angle measurements between two vectors; the cosine of the angle between two vectors is the dot product of the vectors, divided by the product of each of their magnitudes:

$$\displaystyle \cos \theta =\frac{{\text{u}\,\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}};\,\,\,\,\,\,\,\,\,\,\theta ={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}}} \right)$$

(And we don’t need to worry about getting the correct quadrant when putting this in the calculator!)

We might be able to use this formula instead of, say, the Law of Cosines, for applications.

Note that if the dot product of two vectors is 0, the vectors form right angles, or are orthogonal, since the cos of 90° is 0 (and thus the whole expression will be 0).

And remember that we noted above that if two vectors are parallel, then one is a “multiple” of another, or $$\text{u}=a\text{v}$$. For example, the vector $$\text{u}=-2\text{i}+3\text{j}$$ would be parallel to the vector $$\text{v}=-4\text{i}+6\text{j}$$. If vectors are parallel, the angle between them is either 0 (if they are the same vector) or $$\pi$$.

Here are some example problems:

 Angle Between Vector Problems Solutions Find the dot product $$\text{v}\bullet \text{w}$$ and the angle between the two vectors.   Indicate if the vectors are orthogonal, parallel, or neither.      (a) $$\text{u}=4\text{i}-2\text{j};\,\,\,\,\,\text{v}=-2\text{i}+\text{j}$$    (b) $$\text{u}=3\text{i}-\text{j};\,\,\,\,\,\text{v}=-\text{i}+2\text{j}$$    (c) $$\text{u}=-\text{i}+3\text{j};\,\,\,\,\,\text{v}=6\text{i}+2\text{j}$$    (d) $$\text{u}=-\text{i};\,\,\,\,\,\text{v}=-2\text{i}+2\text{j}$$       Picture for (c) above: (a) $$\displaystyle \text{u}\bullet \text{v}=\left( 4 \right)\left( {-2} \right)+\left( {-2} \right)\left( 1 \right)=-10$$ Since $$\displaystyle \left\langle {4,-2} \right\rangle =-2\left\langle {-2,1} \right\rangle$$, the vectors are parallel. \displaystyle \begin{align}\theta &={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| \text{u} \right\|\left\| \text{v} \right\|}}} \right)={{\cos }^{{-1}}}\left( {\frac{{-10}}{{\left( {\sqrt{{{{4}^{2}}+{{{\left( {-2} \right)}}^{2}}}}} \right)\cdot \left( {\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{1}^{2}}}}} \right)}}} \right)\\&={{\cos }^{{-1}}}\left( {-1} \right)=180{}^\circ \end{align}   (b) $$\displaystyle \text{u}\bullet \text{v}=\left( 3 \right)\left( {-1} \right)+\left( {-1} \right)\left( 2 \right)=-5$$ The vectors are neither orthogonal nor parallel. \displaystyle \begin{align}\theta &={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| \text{v} \right\|\left\| \text{w} \right\|}}} \right)={{\cos }^{{-1}}}\left( {\frac{{-5}}{{\left( {\sqrt{{{{3}^{2}}+{{{\left( {-1} \right)}}^{2}}}}} \right)\cdot \left( {\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{2}^{2}}}}} \right)}}} \right)\\&={{\cos }^{{-1}}}\left( {\frac{{-5}}{{\sqrt{{50}}}}} \right)=135{}^\circ \end{align}   (c) $$\displaystyle \text{u}\bullet \text{v}=\left( {-1} \right)\left( 6 \right)+\left( 3 \right)\left( 2 \right)=0$$ Since $$\displaystyle \text{u}\,\bullet \text{v}=0$$, the vectors are orthogonal (see picture to left). \displaystyle \begin{align}\theta &={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| \text{v} \right\|\left\| \text{w} \right\|}}} \right)={{\cos }^{{-1}}}\left( {\frac{0}{{\left( {\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{3}^{2}}}}} \right)\cdot \left( {\sqrt{{{{6}^{2}}+{{2}^{2}}}}} \right)}}} \right)\\&={{\cos }^{{-1}}}\left( {\frac{0}{{20}}} \right)=90{}^\circ \end{align}   (d) $$\displaystyle \text{u}\bullet \text{v}=\left( {-1} \right)\left( {-2} \right)+\left( 0 \right)\left( 2 \right)=2$$ The vectors are neither orthogonal nor parallel. \displaystyle \begin{align}\theta &={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| \text{u} \right\|\left\| \text{v} \right\|}}} \right)={{\cos }^{{-1}}}\left( {\frac{2}{{\left( {\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{0}^{2}}}}} \right)\cdot \left( {\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{2}^{2}}}}} \right)}}} \right)\\&={{\cos }^{{-1}}}\left( {\frac{2}{{\sqrt{8}}}} \right)=45{}^\circ \end{align} Find $$x$$ so that the following vectors are orthogonal:    (a) $$\displaystyle \text{u}=4\text{i}-2\text{j};\,\,\,\,\,\text{v}=x\text{i}+\text{j}$$    (b) $$\text{u}=\text{j};\,\,\,\,\,\,\,\text{v}=3\text{i}+x\text{j}$$ We’ll need to find $$\text{v}\bullet \text{w}$$ and make sure it equals 0: (a) $$\displaystyle \text{u}\bullet \text{v}=\left( 4 \right)\left( x \right)+\left( {-2} \right)\left( 1 \right)=0;\,\,\,\,4x=2;\,\,\,\,x=\frac{1}{2}$$ (b) $$\displaystyle \text{u}\bullet \text{v}=\left( 0 \right)\left( 3 \right)+\left( 1 \right)\left( x \right)=0;\,\,\,\,x=0$$

# 3D Vectors – Vectors in Space

We’ve been dealing with vectors (and everything else!) in the two-dimensional plane, but “real life” is actually three-dimensional, so we need to know how to work in 3D, or space, too.

A 3D coordinate system is typically drawn like this, with the positive $$\boldsymbol{z}$$axis going “up”. Note that the positive $$\boldsymbol{x}$$axis comes forward at you, and the positive $$\boldsymbol{y}$$axis to the right of that, if you’re looking head on. Maybe you can remember this by the expression “Exit. Why?” ($$x$$ – $$z$$ – $$y$$ when looking head on).

Geometric Vectors in 3D are still directed line segments, but in the $$xyz$$-plane. We still can find the vector between two coordinate points by “subtracting” the first vector from the second.

For example, if $$A$$ is $$(-4,2,7)$$ and $$B$$ is $$(-3,8,0)$$, the vector $$\overrightarrow{{AB}}$$ is second point minus first point, or $$\displaystyle \left\langle {{{x}_{2}}} \right.-{{x}_{1}},\,\left. {{{y}_{2}}-{{y}_{1}},\,{{z}_{2}}-{{z}_{1}}} \right\rangle$$ or $$\left\langle {-3-\left( {-4} \right)} \right.,\left. {8-2,0-7} \right\rangle =\left\langle {1,\left. {6,-7} \right\rangle } \right.$$. Note also that vectors can also be written in the form $$\text{ai}+\text{bj}+\text{ck}$$, so this vector can also be written as $$\text{i}+6\text{j}-7\text{k}$$.

The magnitude of the 3D vector, written $$\left\| {AB} \right\|$$ is still the distance between the two points (like taking hypotenuse of a right triangle twice actually), or $$\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}+{{{\left( {{{z}_{2}}-{{z}_{1}}} \right)}}^{2}}}}$$, or with the new vector $$\left\langle {x,} \right.\left. {y,z} \right\rangle$$, it’s just $$\sqrt{{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$$. So, for our points $$A$$ and $$B$$ above, $$\left\| {AB} \right\|=\sqrt{{{{1}^{2}}+{{6}^{2}}+{{{\left( {-7} \right)}}^{2}}+}}=\sqrt{{86}}$$.

## Vector Operations in Three Dimensions

Adding, subtracting 3D vectors, and multiplying 3D vectors by a scalar are done the same way as 2D vectors; you just have to work with three components.

Like for 2D vectors, the dot product of two vectors $$\text{u}=\text{ai}+\text{bj}+\text{ck}$$ and $$\text{v}=\text{di}+\text{ej}+\text{fk}$$ (sort of like multiplying two vectors) is defined as $$\text{u}\bullet \text{v}=\text{ad}+\text{be}+\text{cf}$$; in other words, you multiply the two “$$x$$” parts of the vectors, multiply the two “$$y$$” parts, multiply the two “$$z$$” parts, and then add them together. The result is a scalar (single number).
Again, like for 2D, we use dot products to find the angle measurements between two vectors; the cosine of the angle between two vectors is the dot product of the vectors, divided by the product of each of their magnitudes:

$$\displaystyle \cos \theta =\frac{{\text{u}\,\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}};\,\,\,\,\,\,\,\,\,\,\theta ={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| {\text{u}} \right\|\left\| {\text{v}} \right\|}}} \right)$$

Here are some problems; included is how to get the equation of a sphere:

 3D Vector Problems Solutions For initial point P and terminal point Q, write vector v in from $$\text{ai}+\text{bj}+\text{ck}$$ (its position vector), then find its magnitude:   $$P=\left( {4,-2,0} \right)\,\,\,\,\,\,Q=\left( {0,4,-3} \right)$$ Subtract initial point from terminal point: $$\displaystyle \text{v}=\left\langle {0-4,} \right.\left. {4-\left( {-2} \right),-3-0} \right\rangle =\left\langle {-4,6,-3} \right\rangle =-4\text{i}+6\text{j}-3\text{k}$$   The magnitude of the vector, written $$\left\| {AB} \right\|$$, is the distance between the two points (like the hypotenuse of a right triangle), or $$\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}+{{{\left( {{{z}_{2}}-{{z}_{1}}} \right)}}^{2}}}}$$, or with the new vector $$\left\langle {x,} \right.\left. {y,z} \right\rangle$$, it’s $$\sqrt{{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$$. For points $$A$$ and $$B$$ above, $$\left\| {AB} \right\|=\sqrt{{{{{\left( {-4} \right)}}^{2}}+{{6}^{2}}+{{{\left( {-3} \right)}}^{2}}}}=\sqrt{{61}}$$. Find the distance between P and Q if $$P=\left( {-1,2,5} \right)$$ and $$Q=\left( {0,-5,-3} \right)$$. The distance between two points is similar to the magnitude of a vector: $$\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}+{{{\left( {{{z}_{2}}-{{z}_{1}}} \right)}}^{2}}}}$$. The distance between $$\displaystyle \left( {-1,2,5} \right)$$ and $$\displaystyle \left( {0,-5,-3} \right)$$ is $$\displaystyle \sqrt{{{{{\left( {0-\left( {-1} \right)} \right)}}^{2}}+{{{\left( {-5-2} \right)}}^{2}}+{{{\left( {-3-5} \right)}}^{2}}}}=\sqrt{{114}}$$. Find the equation of the sphere with radius 4 and center $$\left( {2,-1,3} \right)$$. The equation of a sphere is similar to the equation of a circle: $${{\left( {x-a} \right)}^{2}}+{{\left( {y-b} \right)}^{2}}+{{\left( {z-c} \right)}^{2}}={{r}^{2}}$$, where $$\left( {a,b,c} \right)$$ is the center. We have $${{\left( {x-2} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}+{{\left( {z-3} \right)}^{2}}=16$$. Find the dot product $$\text{u}\bullet \text{v}$$ and the angle between the two vectors.   Indicate if the vectors are orthogonal, parallel, or neither.   (a) $$\begin{array}{l}\text{u}=-2\text{i}+4\text{j}+6\text{k}\\\text{v}=\text{i}-2\text{j}-3\text{k}\end{array}$$   (b) $$\begin{array}{l}\text{u}=-\text{i}+4\text{j}+2\text{k}\\\text{v}=2\text{i}-2\text{j}+5\text{k}\end{array}$$ (a) $$\displaystyle \text{u}\bullet \text{v}=\left( {-2} \right)\left( 1 \right)+\left( 4 \right)\left( {-2} \right)+\left( 6 \right)\left( {-3} \right)=-28$$ Since $$\left\langle {-2,4,6} \right\rangle =-2\left\langle {1,-2,-3} \right\rangle$$, the vectors are parallel. $$\displaystyle \theta ={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| \text{u} \right\|\left\| \text{v} \right\|}}} \right)={{\cos }^{{-1}}}\left( {\frac{{-28}}{{\left( {\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{4}^{2}}+{{6}^{2}}}}} \right)\,\cdot \,\left( {\sqrt{{{{1}^{2}}+{{{\left( {-2} \right)}}^{2}}+{{{\left( {-3} \right)}}^{2}}}}} \right)}}} \right)=180{}^\circ$$    (b) $$\displaystyle \text{u}\bullet \text{v}=\left( {-1} \right)\left( 2 \right)+\left( 4 \right)\left( {-2} \right)+\left( 2 \right)\left( 5 \right)=0$$ Since $$\displaystyle \text{u}\bullet \text{v}=0$$, the vectors are orthogonal. $$\displaystyle \theta ={{\cos }^{{-1}}}\left( {\frac{{\text{u}\bullet \text{v}}}{{\left\| \text{u} \right\|\left\| \text{v} \right\|}}} \right)={{\cos }^{{-1}}}\left( {\frac{0}{{\left( {\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{4}^{2}}+{{2}^{2}}}}} \right)\,\cdot \,\left( {\sqrt{{{{2}^{2}}+{{{\left( {-2} \right)}}^{2}}+{{5}^{2}}}}} \right)}}} \right)=90{}^\circ$$

Writing a 3D vector in terms of its magnitude and direction is a little more complicated. Since we can’t really describe a 3D vector in terms of only a magnitude and one direction, we have to get what we call the direction angles:

$$\displaystyle \begin{array}{l}\alpha =\,\text{angle between v and i }\,\text{(positive }x-\text{axis)}\\\beta =\,\text{angle between v and j }\,\text{(positive }y-\text{axis)}\\\gamma =\,\text{angle between v and k }\,\text{(positive }z-\text{axis)}\end{array}$$

Here are what these angles look like:

It turns out for the vector $$\text{v}=\text{ai}+\text{bj}+\text{ck}$$, the direction angles $$\alpha ,\,\beta ,\,\text{and}\,\gamma$$ are:

$$\displaystyle \cos \alpha =\frac{\text{a}}{{\sqrt{{{{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}}}}}=\frac{\text{a}}{{\left\| \text{v} \right\|}},\,\,\,\,\,\,\cos \beta =\frac{\text{b}}{{\sqrt{{{{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}}}}}=\frac{\text{b}}{{\left\| \text{v} \right\|}},\,\,\,\,\,\,\cos \gamma =\frac{\text{c}}{{\sqrt{{{{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}}}}}=\frac{\text{c}}{{\left\| \text{v} \right\|}}$$

It also turns out that $$\displaystyle {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$$

These cosine values are called the direction cosines for the vector v.

To find the 3D vector in terms of its magnitude and direction cosines, we use:

$$\text{v}=\left\| \text{v} \right\|\left( {\left( {\cos \alpha } \right)\text{i}+\left( {\cos \beta } \right)\text{j}+\left( {\cos \gamma } \right)\text{k}} \right)=\left( {\left\| \text{v} \right\|\cos \alpha } \right)\text{i}+\left( {\left\| \text{v} \right\|\cos \beta } \right)\text{j}+\left( {\left\| \text{v} \right\|\cos \gamma } \right)\text{k}$$

Now let’s do a problem:

 3D Vector Problem Solution Find the direction angles of the following vector, and then write the vector in terms of its magnitude and direction cosines:   $$\text{v}=-6\text{i}+3\text{j}-2\text{k}$$ Using the equations above, we see that: \displaystyle \begin{align}\cos \alpha &=\frac{\text{a}}{{\sqrt{{{{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}}}}}=\frac{\text{a}}{{\sqrt{{{{{\left( {-6} \right)}}^{2}}+{{3}^{2}}+{{{\left( {-2} \right)}}^{2}}}}}}=\frac{\text{a}}{{\left\| \text{v} \right\|}}=\frac{{-6}}{7};\,\,\,\,\alpha =149{}^\circ \\\cos \beta &=\frac{\text{b}}{{\sqrt{{{{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}}}}}=\frac{\text{b}}{{\sqrt{{{{{\left( {-6} \right)}}^{2}}+{{3}^{2}}+{{{\left( {-2} \right)}}^{2}}}}}}=\frac{\text{b}}{{\left\| \text{v} \right\|}}=\frac{3}{7};\,\,\,\,\beta =64.6{}^\circ \,\,\,\,\,\\\cos \gamma &=\frac{\text{c}}{{\sqrt{{{{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}}}}}=\frac{\text{c}}{{\sqrt{{{{{\left( {-6} \right)}}^{2}}+{{3}^{2}}+{{{\left( {-2} \right)}}^{2}}}}}}=\frac{\text{c}}{{\left\| \text{v} \right\|}}=\frac{{-2}}{7};\,\,\,\,\gamma =106.6{}^\circ \end{align} Thus, we have: \begin{align}\text{v}&=\left\| \text{v} \right\|\left( {\left( {\cos \alpha } \right)\text{i}+\left( {\cos \beta } \right)\text{j}+\left( {\cos \gamma } \right)\text{k}} \right)\\&=\left( {7\cos 149{}^\circ } \right)\text{i}+\left( {7\cos 64.6{}^\circ } \right)\text{j}+\left( {7\cos 106.6{}^\circ } \right)\text{k}\end{align}

## Cross Products of 3D Vectors

Also, only for vectors 3D vectors, we have what we call a cross product of vectors (also called vector product, since the result is still a vector) of two vectors $$\text{u}={{\text{a}}_{\text{1}}}\text{i}+{{\text{b}}_{\text{1}}}\text{j}+{{\text{c}}_{\text{1}}}\text{k}$$ and $$\text{v}={{\text{a}}_{\text{2}}}\text{i}+{{\text{b}}_{\text{2}}}\text{j}+{{\text{c}}_{\text{2}}}\text{k}$$. The vector that is the cross product of two vectors is actually orthogonal (perpendicular) to both of the original vectors. This is also called the normal vector.

$$\text{u}\times \text{v}=\left( {{{\text{b}}_{\text{1}}}{{\text{c}}_{\text{2}}}-{{\text{b}}_{\text{2}}}{{\text{c}}_{\text{1}}}} \right)\text{i}-\left( {{{\text{a}}_{\text{1}}}{{\text{c}}_{\text{2}}}-{{\text{a}}_{\text{2}}}{{\text{c}}_{\text{1}}}} \right)\text{j}+\left( {{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{2}}}-{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{1}}}} \right)\text{k}$$

This looks crazy! But we can get this cross product using determinants of matrices. We learned about determinants of matrices here in the The Matrix and Solving Systems with Matrices section:

 Matrix Determinant Notes 2 by 2 matrix: With a 2 by 2 matrix, you start with the upper left corner, multiply diagonally down, and then subtract the product where you multiply down diagonally from the upper right corner. 3 by 3 matrix: Multiply each of the top numbers by the determinant of the 2 by 2 matrix that you get by crossing out the other numbers in that top number’s row and column.   For the middle term, you have to subtract.

Here is an example of how we use a determinant to find the cross product of two vectors $$\text{u}=\text{i}+2\text{j}-4\text{k}$$ and $$\text{v}=-\text{i}+5\text{j}+3\text{k}$$:

$$\displaystyle \text{u}\times \text{v}=\left| {\begin{array}{*{20}{c}} \text{i} & \text{j} & \text{k} \\ 1 & 2 & {-4} \\ {-1} & 5 & 3 \end{array}} \right|=\left| {\begin{array}{*{20}{c}} 2 & {-4} \\ 5 & 3 \end{array}} \right|\text{i}-\left| {\begin{array}{*{20}{c}} 1 & {-4} \\ {-1} & 3 \end{array}} \right|\text{j}+\left| {\begin{array}{*{20}{c}} 1 & 2 \\ {-1} & 5 \end{array}} \right|\text{k}\,\,=\,\,\left( {6-\left( {-20} \right)} \right)\text{i}-\left( {3-4} \right)\text{j}+\left( {5-\left( {-2} \right)} \right)\text{k}=26\text{i}+\text{j}+7\text{k}$$

The vector $$\displaystyle 26\text{i}+\text{j}+7\text{k}$$ is orthogonal (perpendicular, normal) to the vectors u and v above.

A few things to remember here. First, we must watch the order of the vectors when we are finding the cross products of vectors; $$\text{u}\times \text{v}$$ is not necessarily the same thing as $$\text{v}\times \text{u}$$.

Also, we can use the right-hand rule to find the direction of the cross product of two vectors by holding up your right hand and make your index finger, middle finger, and thumb all perpendicular to each other (easier said than done!). Then point your index finger in the direction of the first vector (such as u) and your middle finger in the direction of the second vector (such as v). Your thumb will point in the direction of $$\text{u}\times \text{v}$$.

This is something you probably won’t need too much in your math classes, but it can become very handy in Physics. (And remember the directions of 3D vectors as shown in the coordinate system below).

We can use the cross product to find the area of a 3D parallelogram. If that parallelogram has two adjacent sides with vectors u and v, we can take the magnitude of the vectors’ cross product to find its area: $$\left\| {u\times v} \right\|$$. We can also use this if given four vertices of a parallelogram; we would just have to find two adjacent sides of the parallelogram in vector form first.

Here are some cross product problems:

 Cross Product Problem Solution Find a vector orthogonal to both vectors $$\left\langle {4,-2,3} \right\rangle$$ and $$\left\langle {0,-1,2} \right\rangle$$. We can find the cross product of the two vectors and the resulting vector will be the orthogonal vector: \displaystyle \begin{align}\text{u}\,\,\times \text{v}&=\left| {\begin{array}{*{20}{c}} \text{i} & \text{j} & \operatorname{k} \\ 4 & {-2} & 3 \\ 0 & {-1} & 2 \end{array}} \right|=\left| {\begin{array}{*{20}{c}} {-2} & 3 \\ {-1} & 2 \end{array}} \right|\text{i}-\left| {\begin{array}{*{20}{c}} 4 & 3 \\ 0 & 2 \end{array}} \right|\text{j}+\left| {\begin{array}{*{20}{c}} 4 & {-2} \\ 0 & {-1} \end{array}} \right|\text{k}\\&=\left( {-4-\left( {-3} \right)} \right)\text{i}-\left( {8-0} \right)\text{j}+\left( {-4-0} \right)\text{k}\\&=-\text{i}-8\text{j}-4\text{k}\end{align}   The vector orthogonal to both $$\displaystyle \left\langle {4,-2,3} \right\rangle$$ and $$\left\langle {0,-1,2} \right\rangle$$ is $$\displaystyle \left\langle {-1,-8,-4} \right\rangle$$. Find the area of the 3D parallelogram that has two adjacent sides represented by vectors $$\left\langle {4,-2,3} \right\rangle$$ and $$\left\langle {0,-1,2} \right\rangle$$. Since we already found the cross product for this set of vectors in the previous problem, we can just take the magnitude of it to get the area of this parallelogram:   $$\displaystyle \left\| {\left\langle {-1,-8,-4} \right\rangle } \right\|=\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{{\left( {-8} \right)}}^{2}}+{{{\left( {-4} \right)}}^{2}}}}=\sqrt{{81}}=9$$

# The Equation of a Plane

You might also be asked to find the equation of the plane that passes through a given point and is perpendicular to a certain vector, or even the equation of a plane containing three points.

Remember that the equation of a line can be in the standard form $$ax+by=c$$, so the equation of a plane can be in the form $$ax+by+cz=d$$. (These are called Cartesian equations.)

To see what this plane might look like, we can see where it intersects each of the three axes by setting the other variables to 0, for example with the graph $$2x+6y+3z=12$$ (set $$y$$ and $$z$$ equal to 0 and solve for $$x$$, and so on):

As we saw above, the vector that is the cross product of two vectors is actually orthogonal (perpendicular) to both of the original vectors. (This is also called the normal vector.)

It turns out that a vector equation of the plane is:

$$\displaystyle \left\langle {a,b,c} \right\rangle \bullet \left\langle {x-{{x}_{0}},y-{{y}_{0}},z-{{z}_{0}}} \right\rangle =0,\,\,\text{or}\,\,a\left( {x-{{x}_{0}}} \right)+b\left( {y-{{y}_{0}}} \right)+c\left( {z-{{z}_{0}}\,} \right)=0$$,

where $$\left\langle {a,b,c} \right\rangle$$ (also written as $$\displaystyle a\text{i}+b\text{j}+c\text{k}$$) is orthogonal to the plane (the normal vector) and $$\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)$$ is a point on the plane.

This looks really complicated, so let’s do a problem to show it’s not too bad:

 3D Vector Problem Solution Find an equation of the plane that passes through point $$\left( {2,-1,3} \right)$$ and is perpendicular to the vector $$4\text{i}-2\text{j}+3\text{k}$$. From above, we know that the equation of the plane where $$\left\langle {a,b,c} \right\rangle$$ is perpendicular at a certain point $$\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)$$ is $$\displaystyle a\left( {x-{{x}_{0}}} \right)+b\left( {y-{{y}_{0}}} \right)+c\left( {z-{{z}_{0}}\,} \right)=0$$.   For point $$\left( {2,-1,3} \right)$$, use “$$x-2$$”,  “$$y+1$$”, and “$$z-3$$”: $$4\left( {x-2} \right)-2\left( {y+1} \right)+3\left( {z-3} \right)=0$$. Simplify to get $$\displaystyle 4x-2y+3z=19$$. Another way to write the equation we get is to solve for $$z$$: $$\displaystyle z=\frac{{-4x}}{3}+\frac{{2y}}{3}+\frac{{19}}{3}$$.   Note that there’s another way to get “$$d$$” in the plane equation: For vector $$\displaystyle 4\text{i}-2\text{j}+3\text{k}=\left\langle {4,-2,\,3} \right\rangle$$, we have $$4x-2y+3z=d$$. To get $$d$$, we can plug in the point $$\left( {2,-1,3} \right)$$ for $$x$$, $$y$$, and $$z$$: $$d=4\left( 2 \right)-2\left( {-1} \right)+3\left( 3 \right)=19$$. We get the same equation!

To find the equation of the plane containing three points, we first have to find two vectors defined by the points, find the cross product of the two vectors, and then use the Cartesian equation above to find $$d$$:

 3D Vector Problem Solution Find the equation of the plane containing the points $$\text{A}:\left( {4,-1,2} \right)$$, $$\text{B}:\left( {0,-2,-1} \right)$$, and $$\displaystyle \text{C}:\left( {2,0,3} \right)$$. Let’s first find two vectors defined by these points (start from one point and get the vectors that go to the other two points): $$\displaystyle \begin{array}{l}\overrightarrow{{AB}}=\left\langle {0-4,} \right.\left. {-2-\left( {-1} \right),-1-2} \right\rangle =\left\langle {-4,-1,-3} \right\rangle =-4\text{i}-\text{j}-3\text{k}\\\overrightarrow{{AC}}=\left\langle {2-4,} \right.\left. {0-\left( {-1} \right),3-2} \right\rangle =\left\langle {-2,1,1} \right\rangle =-2\text{i}+\text{j}+\text{k}\end{array}$$   We can find the cross product of the two vectors and the resulting vector will be the orthogonal (normal) vector: \displaystyle \begin{align}\overrightarrow{{AB}}\times \overrightarrow{{AC}}&=\left| {\begin{array}{*{20}{c}} \text{i} & \text{j} & \text{k} \\ {-4} & {-1} & {-3} \\ {-2} & 1 & 1 \end{array}} \right|=\left| {\begin{array}{*{20}{c}} {-1} & {-3} \\ 1 & 1 \end{array}} \right|\text{i}-\left| {\begin{array}{*{20}{c}} {-4} & {-3} \\ {-2} & 1 \end{array}} \right|\text{j}+\left| {\begin{array}{*{20}{c}} {-4} & {-1} \\ {-2} & 1 \end{array}} \right|\text{k}\\&=\left( {-1-\left( {-3} \right)} \right)\text{i}-\left( {-4-6} \right)\text{j}+\left( {-4-2} \right)\text{k}\\&=2\text{i}+10\text{j}-6\text{k}\end{align}   Use the point $$\left( {2,0,3} \right)$$: $$2\left( {x-2} \right)+10\left( y \right)-6\left( {z-3} \right)=0$$. The equation of this plane is $$\displaystyle 2x+10y-6z=-14$$.   Another way to write the equation we get is to solve for $$z$$: $$\displaystyle z=\frac{x}{3}+\frac{{5y}}{3}+\frac{7}{3}$$.

We saw that if two planes are perpendicular, the dot product of their normal vectors is 0. If two planes are parallel, their normal vectors are the same, or multiples of each other (with a different “$$d$$”). If two planes are identical, then the whole plane equation (including the “$$d$$”) is the same or a multiple of the other.

Here’s a type of problem you may see:

 Vector Problem Solution Determine if planes are parallel, perpendicular, or identical:   $$\displaystyle {{P}_{1}}=5x-2y+z=6$$ $$\displaystyle {{P}_{2}}=10x-4y+2z=8$$ $$\displaystyle {{P}_{3}}=10x-4y+2z=12$$ $$\displaystyle {{P}_{4}}=2x-y-12z=6$$ Just like with lines, if planes are identical, then one plane equation is a multiple of another. $${{P}_{1}}$$ and $${{P}_{3}}$$ are identical (multiply $${{P}_{1}}$$ by 2).   Just like with lines, if planes are parallel than their normal vectors are proportional (one is multiple of another) so $${{P}_{1}}$$ is parallel to $${{P}_{2}}$$ ($$\left\langle {10,-4,2} \right\rangle$$ is twice $$\left\langle {5,-2,1} \right\rangle$$), but they are not identical since $$6\ne 8$$.   If two planes are perpendicular, the dot product of their normal vectors equals 0, so $${{P}_{1}}$$ and $${{P}_{4}}$$ are perpendicular ($$\left\langle {5,-2,1} \right\rangle \bullet \left\langle {2,-1,-12} \right\rangle =0$$).

# Parametric Form of the Equation of a Line in Space

We can get a vector form of an equation of a line in 3D space by using Parametric Equations (we will go into details about Parametric Equations here).

In two dimensions, we worked with a slope of the line and a point on the line (or the $$y$$-intercept).

In 3D space, we can use a 3D vector $$\left\langle {a,b,c} \right\rangle$$ as the slope of a line, and define that line by an initial point $$\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle$$ (vector that goes through that point and the origin) and the vector  $$\left\langle {a,b,c} \right\rangle$$. Note that the vector $$\left\langle {a,b,c} \right\rangle$$ will be parallel to the line we’re describing, just like a slope going through the origin is parallel to a 2D line. We have three different ways to write this 3D line using parametric equations:

$$\displaystyle \left\langle {x,y,z} \right\rangle =\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle +t\left\langle {a,b,c} \right\rangle$$            $$\displaystyle \begin{array}{c}x={{x}_{0}}+at\\y={{y}_{0}}+bt\\z={{z}_{0}}+ct\end{array}$$            $$\displaystyle \frac{{x-{{x}_{0}}}}{a}=\frac{{y-{{y}_{0}}}}{b}=\frac{{z-{{z}_{0}}}}{c}\,\,\,\,(=t)$$

Notice to get the last form, we solve for $$\boldsymbol {t}$$ in the second set of equations. Also note that to go from the last equation to the first equation, we have to set each to $$t$$, and solve back for $$x$$, $$y$$, and $$z$$.

This looks a little complicated, so let’s do some problems to show it’s not too bad:

 Vector Parametric Problem Solution (a) Find the parametric equation of the line that passes through the points $$\left( {2,-3,1} \right)$$ and $$\left( {4,0,-2} \right)$$. Write down all three forms of this equation.     (b) If this line passes through the XY plane, give the coordinate of the point of intersection. (a) We first need to find the vector v that is parallel to the line that we are trying to find. To get this, we find the vector between the two points: $$\left\langle {4-2} \right.,\left. {0-\left( {-3} \right),-2-1} \right\rangle =\left\langle {2,3,-\left. 3 \right\rangle } \right.$$. To get the vector form of the line, we can use either point as the initial point (we’ll use first one) to get: $$\left\langle 2 \right.,\left. {-3,1} \right\rangle +t\left\langle {2,3,-\left. 3 \right\rangle } \right.$$, or  $$\left\langle 2 \right.+2t,\left. {-3+3t,1-3t} \right\rangle$$.   The other forms are $$\displaystyle \begin{array}{c}x=2+2t\\y=-3+3t\\z=1-3t\end{array}$$ and (solving for $$t$$): $$\displaystyle \frac{{x-2}}{2}=\frac{{y+3}}{3}=\frac{{z-1}}{{-3}}\,\,\left( {\frac{{1-z}}{3}} \right)$$.   (b) To get the point where this line passes through the XY plane, we can set $$z$$ to 0 to get $$t$$ in this equation: $$\displaystyle z=1-3t;\,\,\,0=1-3t;\,\,\,\,t=\frac{1}{3}$$, and then put this value of $$t$$ in the line’s equation: $$\displaystyle \left\langle 2 \right.,\left. {-3,1} \right\rangle +t\left\langle {2,3,-\left. 3 \right\rangle } \right.;\,\,\,\,\left\langle 2 \right.,\left. {-3,1} \right\rangle +\left( {\frac{1}{3}} \right)\left\langle {2,3,-\left. 3 \right\rangle } \right. =\left\langle {2\frac{2}{3},-2,0} \right\rangle$$. The point of intersection is $$\displaystyle \left( {2\frac{2}{3},-2,0} \right)$$. Find the equation of the plane that passes through the point $$\left( {3,-3,1} \right)$$, and is perpendicular to the line $$\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{2}$$.       (Note that $$\displaystyle \frac{{2-x}}{3}=\frac{{x-2}}{{-3}}$$) We know that the vector equation of the plane is: $$\displaystyle \left\langle {a,b,c} \right\rangle \bullet \left\langle {x-{{x}_{0}},y-{{y}_{0}},z-{{z}_{0}}} \right\rangle =0,\,\,\text{or}\,\,a\left( {x-{{x}_{0}}} \right)+b\left( {y-{{y}_{0}}} \right)+c\left( {z-{{z}_{0}}} \right)=0$$, where $$\left\langle {a,b,c} \right\rangle$$ is orthogonal to the plane (the normal vector) and $$\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)$$ is a point on the plane. Since we are finding the equation of the plane that is perpendicular to the line and not containing the line, we don’t have to take a cross product.   The direction vector of the given line is $$\left\langle {-3} \right.,\left. {4,2} \right\rangle$$ (to get this, set each expression to $$t$$, and solve $$x$$, $$y$$, and $$z$$, multiplying the $$x$$ expression by –1). To get the equation of the plane that is perpendicular to the vector $$\left\langle {-3} \right.,\left. {4,2} \right\rangle$$, we have $$-3x+4y+2z=d$$.   Use the point $$\left( {3,-3,1} \right)$$: $$-3\left( {x-3} \right)+4\left( {y+3} \right)+2\left( {z-1} \right)=0;\,-9x+4y+2z=-19$$, or $$\displaystyle z=\frac{3}{2}x-2y-\frac{{19}}{2}$$. Find the equation of the plane that passes through the point $$\left( {3,-3,1} \right)$$, and contains the line $$\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{2}$$.       (Note that $$\displaystyle \frac{{2-x}}{3}=\frac{{x-2}}{{-3}}$$) For this problem, we will need to use the cross product, since the equation of the plane uses the vector that is normal to the plane, and the cross section gives us that. We know the point $$\left( {3,-3,1} \right)$$ is on the plane (given). We can get two more points on this plane by setting each expression of the line equation to $$t$$, solving for $$x$$, $$y$$, and $$z$$: $$\displaystyle (x=-3t+2;\,\,y=4t-1;\,\,z=2t)$$, then using different values for $$t$$.   For $$t=0$$, we get the point $$\left( {2,-1,0} \right)$$ and when $$t=1$$, we get the point $$\left( {-1,3,2} \right)$$. Now we can use the method above (3D Vector Problem) to find the equation of the plane, given three points.   The vector between first two points is $$\displaystyle -1\text{i}+2\text{j}-\text{k}$$, and vector between the last two points is $$\displaystyle -3\text{i}+4\text{j}+2\text{k}$$. We can find the cross product of the two vectors and the resulting vector will be the orthogonal (normal) vector: $$\displaystyle \left| {\begin{array}{*{20}{c}} \text{i} & \text{j} & \text{k} \\ {-1} & 2 & {-1} \\ {-3} & 4 & 2 \end{array}} \right|=\left| {\begin{array}{*{20}{c}} 2 & {-1} \\ 4 & 2 \end{array}} \right|\text{i}-\left| {\begin{array}{*{20}{c}} {-1} & {-1} \\ {-3} & 2 \end{array}} \right|\text{j}+\left| {\begin{array}{*{20}{c}} {-1} & 2 \\ {-3} & 4 \end{array}} \right|\text{k}=\,8\text{i}+5\text{j}+2\text{k}$$   Use the point $$\left( {2,-1,0} \right)$$): $$8\left( {x-2} \right)+5\left( {y+1} \right)+2\left( z \right)=0;\,\,\,8x+5y+2z=11$$, or $$\displaystyle z=-4x-\frac{{5y}}{2}+\frac{{11}}{2}$$. Find the equation of the plane that contains the parallel lines $$\displaystyle x+1=\frac{{y-2}}{2}=\frac{{z-4}}{{-1}}$$ and $$\displaystyle \frac{{x-2}}{{-1}}=\frac{{y-3}}{{-2}}=z$$.   (Note that only parallel lines and intersecting lines have a plane containing the lines. Skew lines do not.) We can see that the lines are parallel since their directional vectors are multiples of each other (look at denominators), and the lines are different.   We can take the cross product of one of these directional vectors and also another vector containing any point on the first line and any point on the second line (this way the cross-product equation will work). This will give us a vector perpendicular or normal to the plane. We’ll use points $$\left( {-1,2,4} \right)$$ and $$\left( {2,3,0} \right)$$ to get $$\left\langle {2-\left( {-1} \right)} \right.,\left. {3-2,0-4} \right\rangle =\left\langle {3,1,-\left. 4 \right\rangle } \right.$$, and also $$\left\langle {1,2,-\left. 1 \right\rangle } \right.$$, the directional vector of the first line: $$\displaystyle \begin{array}{l}\text{u}\times \text{v}\,=\left| {\begin{array}{*{20}{c}} \text{i} & \text{j} & \text{k} \\ 3 & 1 & {-4} \\ 1 & 2 & {-1} \end{array}} \right|\,\,=\,\,\left| {\begin{array}{*{20}{c}} 1 & {-4} \\ 2 & {-1} \end{array}} \right|\text{i}-\left| {\begin{array}{*{20}{c}} 3 & {-4} \\ 1 & {-1} \end{array}} \right|\text{j}+\left| {\begin{array}{*{20}{c}} 3 & 1 \\ 1 & 2 \end{array}} \right|\text{k}\\=7\text{i}-\text{j}+5\text{k}\end{array}$$   Use the point $$\left( {2,3,0} \right)$$: $$7\left( {x-2} \right)-1\left( {y-3} \right)+5\left( z \right)=0;\,\,\,7x-y+5z=11$$, or $$\displaystyle z=\frac{{-7x}}{5}+\frac{y}{5}+\frac{{11}}{5}$$.

Here are some more complicated problems; these can get a little tricky!

 Vector Parametric Problem Solution The plane $$\displaystyle 5x-2y+z=6$$ contains the line $$\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{k}$$.   Find $$k$$. We can get a vector that is normal (perpendicular) to the given plane $$\displaystyle 5x-2y+z=6$$; this is$$\displaystyle \left\langle {5,-2,1} \right\rangle$$ .   We can also find a vector parallel to the given line $$\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{k}$$; this is $$\displaystyle \left\langle {-3,4,k} \right\rangle$$. (To do this, we rewrite the equations in vector form: $$\displaystyle \left\langle {x,y,z} \right\rangle =\left\langle {-2,1,0} \right\rangle +t\left\langle {-3,4,k} \right\rangle$$, noting that we had to multiply the $$x$$ expression by –1).   If these two vectors are perpendicular, then we know that the plane contains the line. Take the cross product of the two vectors, set it to 0, and solve for $$k$$: $$\displaystyle \left\langle {-3,4,k} \right\rangle \bullet \,\,\left\langle {5,-2,1} \right\rangle =0$$, $$\displaystyle -3\cdot 5+4\cdot -2+k\cdot 1=0$$ or $$\displaystyle -3\cdot 5+4\cdot -2+k\cdot 1=0$$. Solving, we get $$k=23$$. The line $$\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{2}$$ and plane $$\displaystyle 5x-2y+z=30$$ intersect at point $$Q$$.   Find $$Q$$. Rewrite the line $$\displaystyle \frac{{2-x}}{3}=\frac{{y+1}}{4}=\frac{z}{2}$$ in vector parametric form: $$\displaystyle \left\langle {x,y,z} \right\rangle =\left\langle {-2,1,0} \right\rangle +t\left\langle {-3,4,2} \right\rangle$$, or $$\displaystyle \begin{array}{l}x=-2-3t\\y=1+4t\\z=2t\end{array}$$.   We can now substitute $$x$$, $$y$$, and $$z$$ in the plane equation $$\displaystyle 5x-2y+z=30$$, since we want the point of intersection: $$\displaystyle 5\left( {-2-3t} \right)-2\left( {1+4t} \right)+\left( {2t} \right)=30$$. Solving, we get $$t=-2$$.   We want the point $$Q$$ of this intersection, so we have $$\displaystyle \left\langle {-2,1,0} \right\rangle +t\left\langle {-3,4,2} \right\rangle =\left\langle {-2,1,0} \right\rangle +\left( {-2} \right)\left\langle {-3,4,2} \right\rangle =\left( {-4,-7,-4} \right)$$. The point $$P$$ is at the foot of the perpendicular line from the point $$\left( {3,1,0} \right)$$ to the plane $$\displaystyle 5x-2y+z=73$$.   Find $$P$$. We can get a vector that is normal (perpendicular) to the plane $$\displaystyle 5x-2y+z=73$$; this is $$\displaystyle \left\langle {5,-2,1} \right\rangle$$. Since the perpendicular line (where $$P$$ is at the foot) goes through the point $$\left( {3,1,0} \right)$$ and is normal (perpendicular) to the plane, this line’s vector form is $$\displaystyle \left\langle {x,y,z} \right\rangle =\left\langle {3,1,0} \right\rangle +t\left\langle {5,-2,1} \right\rangle$$,  or $$\displaystyle \begin{array}{l}x=3+5t\\y=1-2t\\z=t\end{array}$$.   We can now substitute $$x$$, $$y$$, and $$z$$ in the plane equation $$\displaystyle 5x-2y+z=6$$, since we want the point of intersection: $$\displaystyle 5\left( {3+5t} \right)-2\left( {1-2t} \right)+\left( t \right)=73$$   Solving, we get $$t=2$$. We want the point $$P$$ of this intersection, so we have $$\displaystyle \left\langle {3,1,0} \right\rangle +t\left\langle {5,-2,1} \right\rangle =\left\langle {3,1,0} \right\rangle +\left( 2 \right)\left\langle {5,-2,1} \right\rangle =\left( {13,-3,2} \right)$$.

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