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Distance (2, 0, -3) (4, 3, 1)

Step by step, with the reason for every step · Points and distance in space · Calculus 3

Step by step

The question

Question: find the distance between (2, 0, −3) and (4, 3, 1).

(2, 0, −3) and (4, 3, 1)

1 Plot the points

Draw the three axes: x toward us, y right, z up.

Point A: (2, 0, −3).

Point B: (4, 3, 1).

The distance is the length of this straight segment.

2 The differences

Along x: x goes from 2 to 4.

4 minus 2 is 2.

4 − 2 = 2

Along y: y goes from 0 to 3.

3 minus 0 is 3.

3 − 0 = 3

Up z: z goes from −3 to 1.

1 minus (−3) is 4.

1 − (−3) = 4

3 Square and add

On the flat, the x and y moves make a right triangle.

Its long side is this green line.

The green line and the move up z make a second right triangle.

Its long side is AB.

So Pythagoras works twice.

Square the three differences, add them, then take the square root.

d = √22 + 32 + 42

2 squared is 4.

3 squared is 9.

4 squared is 16.

d = √4 + 9 + 16

4 Take the square root

Add: 4 plus 9 plus 16 is 29.

d = √29

Careful!

29 is d squared.

We still take the square root.

Careful: not d = 29 ✗

29 is not a square number, so √29 is the exact answer.

d = √29

With a calculator, that is about 5.4, to the nearest tenth.

d ≈ 5.4

5 Check

Let's check: square the distance.

We get 29, the same as 4 plus 9 plus 16.

Right!

d² = 29 and 4 + 9 + 16 = 29

Going from B to A instead only changes the signs.

The squares are the same.

Right!

(2 − 4)² = 4

Let's look back.

First: plot the points.

Then: the differences.

Then: square and add.

Then: take the square root.

Last: check.

Answer
d ≈ 5.4

Your turn

Try this one yourself, then check:

Find the distance between (2, 0, −3) and (5, 3, 2)
Show the answer
d = √43 ≈ 6.6

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