Calculus 3 Tutor, Volume I Worksheet 1 3D Coordinates · 2019-08-20 · Calculus 3 Tutor, Volume I Worksheet 1 3D Coordinates. Worksheet for Calculus 3 Tutor, Volume I, Section 1:

Calculus 3 Tutor, Volume I

Worksheet 1

3D Coordinates

Worksheet for Calculus 3 Tutor, Volume I, Section 1:

3D Coordinates

1. Find the (x, y, z) coordinates for the following points plotted in the 3-dimensional

plane.

(a) See diagram:

Plot for 1a

c©2019 MathTutorDVD.com 1

(b) See diagram:

Plot for 1b


(c) See diagram:

Plot for 1c


(d) See diagram:

Plot for 1d


(e) See diagram:

Plot for 1e


(f) See diagram:

Plot for 1f


(g) See diagram:

Plot for 1g


(h) See diagram:

Plot for 1h


2. Plot the following points:

(a) (5, 0, 2);

Plot for 2a


(b) (−3, 4, 0);

Plot for 2b


(c) (0,−4,−2);

Plot for 2c


(d) (6, 1, 5);

Plot for 2d


(e) (−3, 2, 3);

Plot for 2e


(f) (2, 0, 0);

Plot for 2f


(g) (6,−5, 4);

Plot for 2g


(h) (6, 5,−4);

Plot for 2h


3. The main purpose of this lesson is to become familiar with the 3D plane and how it is

similar to the 2D plane. In the 2D plane, a point (x, y) falls in one of four quadrants.

In the 3D plane, a point (x, y, z) falls in one of eight octants.

(a) In the 2D plane, a point is in the first quadrant if x and y are both positive; the

second, if x is negative but y is positive; the third, if x and y are both negative;

and the fourth, if x is positive but y is negative. Label all four quadrants in the

following diagram. The x, y plane is the flat plane where z is zero.

Plot for 3a


(b) In the 3D plane, a point is in octants 1-4 if z is positive. If z is positive, then the

octant number of (x, y, z) is the same as the quadrant number of (x, y). Label the

first four octants on the following diagram.

Plot for 3b


(c) In the 3D plane, a point is in octants 5-8 if z is negative. If z is negative, then the

octant number of (x, y, z) is four plus the quadrant number of (x, y). Label octants

five through eight on the following diagram.

Plot for 3c

(d) Determine the octant number of the following points:

i. (1, 1, 5);

ii. (−2, 3, 1);

iii. (3, 2,−1);

iv. (−4, 2,−3);


v. (−10,−20,−30);

vi. (−5,−5, 10);

vii. (1,−1,−1);

viii. (100,−10, 1);

4. Find the distance between the following points in the 3D Cartesian plane:

(a) (0, 0, 0) and (−8, 0, 15);

(b) (0, 1, 0) and (−4, 2, 2√

2);


(c) (−1, 3, 5) and (−5,−1, 7);

(d) (2,−2, 16) and (−4,−11, 18);

(e) (5, 4,−7) and (1,−6,−3);

(f) (−4,−3, 2) and (6, 7, 1).


5. For the point P = (3, 6, 2):

(a) Find the distance from P to the origin by finding the length of both brown lines,

and computing the length of the hypotenuse of the right triangle formed by the

two brown lines. Note that the brown lines intersect where z = 0.

Plot for 5a


(b) Find the distance from P to the origin by finding the length of both blue lines, and

computing the length of the hypotenuse of the right triangle the blue lines form.

Plot for 5b


(c) Find the distance from P to the origin by finding the length of both green lines,

and computing the length of the hypotenuse of the right triangle the green lines

form.

Plot for 5c


(d) Find the distance from P to the origin using the distance formula in 3D coordi-

nates.

6. Prove the 3D distance formula by using the Pythagorean theorem as above:

(a) Prove that the distance between (x, y, z) and (0, 0, 0) is√x2 + y2 + z2.

(b) Prove that the distance between (x1, y1, z1) and (x2, y2, z2) is√(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2


Answer key.

1. Identifying points plotted in the 3D plane:

1(a). Answer: (3, 2, 0) . The point is 3 units away from the origin along the x-axis and

2 units away from the origin along the y-axis. Both the 3 and the 2 are counted in

the positive direction, since the point is in the positive quadrant (with the labeled

x and y axes). The point is in the (x, y) plane; it doesn’t have any height relative

to that plane, so z = 0. Then, the point is (3, 2, 0) .

1(b). Answer: (−3, 5, 0) . The point is 3 units away from the origin along the x-axis.

Then, the x-coordinate is −3, since it is in the negative direction along the x-axis

(that is, away from the arrow in x). We can count that the point is 5 units away

from the origin along the y-axis; that is, there are five notches along the y-axis

until we get to the level of the point. Finally, the point is in the (x, y) plane, so

z = 0. Then the coordinates of the point are (−3, 5, 0) .


1(c). Answer: (3,−2, 1) . The point has a positive x-coordinate, and we can count

the notches to find that x = 3. We can also count in y to find that y = −2. The

base of the red “height” line is at (3,−2), and the “height” line is up to height 1.

Therefore, the coordinates of the point are (3,−2, 1) .

1(d). Answer: (8, 3,−3) . Counting the notches in x, we find that x = 8. Then, from

(8, 0, 0), we can count the notches in y to find that y = 3. Finally, z is 3 units below

the surface, so z = −3. Then, the point is (8, 3,−3) .


1(e). Answer: (−3, 0,−2) . Since the point is directly below the x-axis, we have that

y = 0. It is on the third x-marker behind the origin, so x = −3. Finally, the height

marker is two units below the origin. This means that z = −2 so the point is

(−3, 0,−2) .

1(f). Answer: (0, 0, 6) . Since the point is directly above the origin, x = y = 0. By

counting the notches, we can tell that the height is z = 6. Then, the point is

(0, 0, 6) .


1(g). Answer: (−2,−4,−3) . The point is “behind” the origin along both the x and y

axis, so x and y are both negative. Counting, we see that x = −2 and y = −4.

Likewise, the point is under the plane z = 0, so z is negative and equal to −3, so

the point’s coordinates are (−2,−4,−3) .

1(h). Answer: (8, 0, 6) . The point is above the x-axis, so y = 0. It is above the point

(8, 0), where the red height line intersects the axis. The red height line is six units

long as marked on the z-axis, so the point is (8, 0, 6) .


2. Plotting points given their coordinates:

2(a). To plot (5, 0, 2), we mark 5 units along the x-axis where y is zero. Then we draw

the red line for height two units high in the upwards direction to get the following

point:

Plot for 2a


2(b). To plot (−3, 4, 0), we plot (−3, 4) on the (x, y) grid, since z = 0:

Plot for 2b


2(c). To plot (0,−4,−2), we plot (0,−4) on the plane. Here x is zero, so the point is

below the y-axis. It is below (0,−4), and we plot (0,−4,−2) as 2 units below

(0,−4):

Plot for 2c


2(d). To plot (6, 1, 5), we find (6, 1) on the (x, y) grid. That is the start of the red height

segment, which is 5 units tall (corresponding to 5 on the z-axis). Then the point

is:

Plot for 2d


2(e). We plot (−3, 2) and then draw a height segment from that point that appears to

have length 3 to get to (−3, 2, 3):

Plot for 2e


2(f). The point (2, 0, 0) is in the (x, y) plane, since the z-coordinate is zero. It is on the

positive x-axis, since x is positive and y is zero:

Plot for 2f


2(g). The point (6,−5, 4) is 4 units above the point (6,−5), where we can start to draw

the height segment. Then the point is at:

Plot for 2g


2(h). The point (6, 5,−4) is 4 units below the point (6, 5), from which we can “sink” the

height segment of length 4 (corresponding to −4 on the z-axis) to plot the point

as follows:

Plot for 2h


3. Exploring octants:

3(a). The 2D plane, the (x, y) plane, is demarcated by gridlines. Here, the first quadrant

is where x and y are both positive; that is, the quadrant with the darker lines on

both sides. The second quadrant is where x is negative but y is positive; that is,

where θ is between π2

and π. This quadrant is on the same side of the x-axis as

the first quadrant, but on the opposite side of the y-axis. The third quadrant is

diagonally opposite from the first quadrant. Finally, the fourth quadrant is where x

is positive but y is negative; that is, where θ is between 3π2

and 2π. This quadrant

is on the same side of the y-axis as the first quadrant, but on the opposite side of

the x-axis. The plot is as follows:

Plot for 3a

Another way to check these results is to recall that, in the 2D Cartesian plane,

the quadrants are ordered counterclockwise. That is, after I, the next quadrant

counterclockwise is II, then III, and then IV. In this diagram, the quadrants are

also ordered counterclockwise.


3(b). The first four octants are in the same positions as the first four quadrants. That

is, the first octant is the region partitioned by the blue planes above the first

quadrant; the second octant is the region partitioned by the blue planes above

the second quadrant; the third octant is the region partitioned by the blue planes

above the third quadrant; and the fourth octant is the region partitioned by the

blue planes above the fourth quadrant. The plot is as follows:

Plot for 3b


3(c). The next four octants, with z negative, are the same as the first four octants but

with an offset of four. That is, the region below the first octant/first quadrant is

the fifth octant; the region below the second octant/second quadrant is the sixth

octant; the region below the third octant/third quadrant is the seventh octant; and

the region below the fourth octant/fourth quadrant is the eighth octant:

Plot for 3c


3(d). Sorting points into octants:

3(d)1. Answer: this point is in Octant I . Since all three coordinates are positive,

it makes sense that it would be in the first octant. It is above the point (1, 1),

which is in the first quadrant; since z is positive, it is also in the first octant.

3(d)2. Answer: this point is in Octant II . The point’s projection onto the (x, y)

plane, (−2, 3), is in the second quadrant. Since z is positive, the octant num-

ber matches the quadrant number, so this point is in the second octant.

3(d)3. Answer: this point is in Octant V . The projection onto the (x, y) plane,

(3, 2), is in the first quadrant. Therefore, (3, 2, 1) would be in the first octant.

However, since z is negative, we need to add 4 to the octant number. Then

the point is in the fifth octant.


3(d)4. Answer: this point is in Octant VI . The point (−4, 2) is in the second quad-

rant, since x is negative but y is positive. However, since the 3D point is below

the (x, y) plane, the octant number is six instead of two.

3(d)5. Answer: this point is in Octant VII . Since it is below the third quadrant

(since all three coordinates are negative), the octant number is 3 + 4 = 7.

3(d)6. Answer: this point is in Octant III . This point is above (−5,−5), which is in

the third quadrant. Therefore, the octant number is also three.


3(d)7. Answer: this point is in Octant VIII . This point is below (1,−1), which is in

the fourth quadrant since x is positive but y is negative. Then, since it is below

z = 0, the octant number is eight.

3(d)8. Answer: this point is in Octant IV . It is above (100,−10), which is in the

fourth quadrant. Therefore the octant number is also four.


4. Finding distance in the 3D plane:

4(a). Answer: 17 . The distance between (x1, y1, z1) and (x2, y2, z2) is

D =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

Here,

(x1, y1, z1) = (0, 0, 0)

and

(x2, y2, z2) = (−8, 0, 15)

Then, the distance is√(−8− 0)2 + (0− 0)2 + (15− 0)2 =

√82 + 152 =

√64 + 225 =

√289

which is 17 . Note that distance must always be positive, so there is no ambiguity

about the sign of the square root.


4(b). Answer: 5 . The distance between (x1, y1, z1) and (x2, y2, z2) is

D =

√(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2

Here,

(x1, y1, z1) = (0, 1, 0)

and

(x2, y2, z2) = (−4, 2, 2√

2)

Then,

D =

√(−4− 0)2 + (2− 1)2 +

(2√

2− 0)2

=

√42 + 12 +

(2√

2)2

=√

16 + 1 + 8

which evaluates to√

25 = 5

so the answer is 5 .


4(c). Answer: 6 . We compute:

D =

√(−1− (−5))2 + (3− (−1))2 + (5− 7)2

This is

D =√

42 + 42 + 22 =√

16 + 16 + 4 =√

36

which evaluates to 6 .

4(d). Answer: 11 . We have that

∆x = 2− (−4) = 6

∆y = −2− (−11) = 9

∆z = 16− 18 = −2

Then,

D =

√(∆x)2 + (∆y)2 + (∆z)2 =

√62 + 92 + 22 =

√36 + 81 + 4

which is√

121, which is 11 .


4(e). Answer: 2√

33 . The distance is

D =

√(5− 1)2 + (4− (−6))2 + (−7− (−3))2 =

√42 + 102 + (−4)2

This is

D =√

132

which reduces to 2√

33 .

4(f). Answer:√

201 . We compute

D =√

(6− (−4))2 + (7− (−3))2 + (1− 2)2 =√

102 + 102 + 12

which is√

201 .


5. Calculating the same distance in different ways:

5(a). Answer: the red lengths are 3√

5 and 2 , and the distance from P to the origin

is 7 . In the (x, y) plane, the origin is (0, 0), and the end of the longer brown line

is (3, 6). Then, the length of the longer brown line is√(3)2 + (6)2 =

√9 + 36 =

√45

which is 3√

5 . The shorter brown line is the distance between (3, 6, 0) and

(3, 6, 2), which is 2 . Then, the distance from P to the origin is the hypotenuse of

right triangle OQP where O is the origin and Q is the right angle at (3, 6, 0) where

the brown lines intersect. We have found that OQ is 3√

5 and QP is 2, so the

hypotenuse is √(3√

5)2

+ 22 =√

45 + 4 =√

49

which is 7 .

Plot for 5a


5(b). Answer: the blue lengths are√

13 and 6 , and the distance from P to the origin

is 7 . The shorter blue line is the distance from (0, 0, 0) to (3, 0, 2). In the xz-plane,

this is the distance from (0, 0) to (3, 2), which is

√32 + 22 =

√13

The long blue line is the distance from (3, 0, 2) to (3, 6, 2), which is 6 . Then, these

blue sides are the legs of a right triangle, and the hypotenuse of this right triangle

is the distance from P to the origin. By the Pythagorean theorem, the hypotenuse

is √(√13)2

+ 62 =√

13 + 36 =√

49

which is 7 .


5(c). Answer: the green lengths are 2√

10 and 3 , and the distance from P to the

origin is 7 . The longer green line is the distance from (0, 0, 0) to (0, 6, 2). The

other endpoint is (0, 6, 2) since two of its coordinates are the same as those of P ,

but the x coordinate is equal to zero. This distance is

√62 + 22 =

√36 + 4 =

√40

which reduces to 2√

10 . The shorter green line, from (0, 6, 2) to (3, 6, 2), has

length 3 . The green sides are the legs of a right triangle, and the hypotenuse

of this right triangle is the distance from P to the origin. By the Pythagorean

theorem, the hypotenuse is√(2√

10)2

+ 32 =√

40 + 9 =√

49

which is 7 .


5(d). Answer: 7 . By the 3D distance formula, the distance from P to the origin is

√62 + 32 + 22 =

√36 + 9 + 4 =

√49

which is 7 , as we have already found in a number of different ways.


6. Proving the 3D distance formula:

6(a). Answer: D =√x2 + y2 + z2 . The distance between (0, 0, 0) and (x, y, 0) is√

x2 + y2

Then, there is a right triangle formed with the three vertices O = (0, 0, 0); Q =

(x, y, 0); and P = (x, y, z). The length of the side OQ between (0, 0, 0) and (x, y, 0)

is√x2 + y2, and the length of the side QP between (x, y, 0) and (x, y, z) is z.

This triangle is shown in the following diagram. It is a right triangle, because the

vertical height is perpendicular to the flat plane.

Plot for 6a

Then the hypotenuse OP , which is the distance between (x, y, z) and the origin,

has length

D =

√(√x2 + y2

)2+ (z)2

which is D =√x2 + y2 + z2 .


6(b). Answer: D =

√(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 . The distance between (x1, y1, z1)

and (x2, y2, z1) is the distance between (x1, y1) and (x2, y2), since z is the same.

This distance is √(x1 − x2)2 + (y1 − y2)2

Then, there is a right triangle formed with the three vertices (x1, y1, z1); (x2, y2, z1);

and (x2, y2, z2). The first side between (x1, y1, z1) and (x2, y2, z1) has length√(x1 − x2)2 + (y1 − y2)2

The second side, between (x2, y2, z1) and (x2, y2, z2), is

|z1 − z2|

units long, since distance cannot be negative. Then, the hypotenuse is√(√(x1 − x2)2 + (y1 − y2)2

)2

+ (|z1 − z2|)2

The absolute value disappears since the square will always be positive. Then,

the distance is D =

√(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2 .


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Calculus 3 Tutor, Volume I Worksheet 1 3D Coordinates · 2019-08-20 · Calculus 3 Tutor, Volume I Worksheet 1 3D Coordinates. Worksheet for Calculus 3 Tutor, Volume I, Section 1: