A third axis
In space, a point needs three coordinates. Two axes, x and y, lie flat at right angles to each other, and a third, the z-axis, stands at right angles to both of them. The point (2, 3, 6) is 2 along the x-axis, 3 along the y-axis and 6 up the z-axis from the origin.
A vector in three dimensions has three components in the same way. The vector (2, 3, 6) is a movement of 2 in the x-direction, 3 in the y-direction and 6 in the z-direction.
Make those three movements the edges of a box, 2 by 3 by 6. Starting at one corner of the box, the vector (2, 3, 6) ends at the opposite corner, so it is the diagonal that runs through the inside of the box, called the space diagonal.
A box 2 by 3 by 6. The vector (2, 3, 6) runs from one bottom corner to the opposite top corner.
The length: Pythagoras twice
The length of (2, 3, 6) is the length of that space diagonal, and it takes Pythagoras twice.
First, across the floor of the box. The floor is 2 by 3, so its diagonal is .
Second, up from the end of that diagonal. The floor diagonal, the upright edge of 6 and the space diagonal make a right triangle standing inside the box, with the right angle at the bottom of the edge. The space diagonal is its hypotenuse, so its length is .
The square root and the square of cancel, so the three squares end up under one root: . In general, , one more square than in two dimensions.
The floor diagonal , the upright edge 6 and the space diagonal 7 make the shaded right triangle: .
the floor diagonal is √(x² + y²) = 3.61; stand z = 6 on its end and Pythagoras again gives |v| = √(3.61² + 6²) = 7, which is √(x² + y² + z²)
Make the box 3 by 4 by 12 and read the diagonal
The same 2 by 3 by 6 box. The gold floor diagonal is to 2 decimal places, and the space diagonal is . Drag the ends of the three edges to change the box: make it 3 by 4 by 12, and the floor diagonal is 5 and the space diagonal 13.
i, j and k
In two dimensions, i = (1, 0) is one step along the x-axis and j = (0, 1) one step along the y-axis. In three dimensions they become i = (1, 0, 0) and j = (0, 1, 0), and a third unit vector joins them: k = (0, 0, 1), one step along the z-axis. Each of the three has length 1.
So (2, 3, 6) = 2i + 3j + 6k: 2 steps of i, 3 of j and 6 of k. A component of 0 leaves its letter out, so (4, 0, −1) = 4i − k.
Adding, subtracting and scaling
Vectors in three dimensions are added one place at a time, exactly as in two: (1, 2, 2) + (2, 1, 4) = (1 + 2, 2 + 1, 2 + 4) = (3, 3, 6). The third places add just like the first two.
Subtraction and multiplying by a scalar work the same way: (2, 3, 6) − (1, −2, 4) = (2 − 1, 3 − (−2), 6 − 4) = (1, 5, 2), and 2(1, 2, 2) = (2, 4, 4).
The rules carry over because the three axes are at right angles to each other. A movement along one axis leaves the other two coordinates unchanged, so each place can be worked on its own.
The dot product and the unit vector
The dot product gains a third product: (2, 3, 6) · (1, −2, 4) = 2 × 1 + 3 × (−2) + 6 × 4 = 2 − 6 + 24 = 20.
The unit vector divides each component by the length: (2, 3, 6) has length 7, so its unit vector is . Check: .
The usual mistakes
Adding the components. 2 + 3 + 6 = 11 is the length of a walk along three edges of the box; the straight line through the inside is shorter, 7.
Leaving out the square root. 4 + 9 + 36 = 49 is the length squared; the length is .
Dropping the third component in a sum. (1, 2, 2) + (2, 1, 4) is (3, 3, 6), not (3, 3, 2).
Multiplying the components in a sum. Adding two movements makes one longer movement; (2, 2, 8) multiplies the places instead of adding them.
A cable across a warehouse
In the application below, the axes run along three edges of a room from one corner on the floor, so the cable to the far top corner is a space diagonal: its length is the square root of three squares. The hook at the middle of the cable is at the mean of each coordinate, halfway along, halfway across and halfway up.
Worked example: A Cable Across a Warehouse from a Floor Corner to the Far Top Corner
Question A warehouse is a cuboid 12 m long, 4 m wide and 3 m high. With axes along its edges from the corner O on the floor, a cable runs straight from O to a light L at (12, 4, 3) in the opposite top corner. A sensor S is at (0, 4, 3). (a) Find the length of the cable. (b) A hook holds the cable at its midpoint M. Find the coordinates of M and the distance from the sensor to the hook.
1.The cable is the vector OL = 1243: along the length, across the width, and up the height of the room.
The cable is the vector OL = 1243: along the length, across the width and up the height of the room. 2.(a) In three dimensions the length is √122 + 42 + 32 = √144 + 16 + 9 = √169 = 13 m.
(a) Its length is √122 + 42 + 32 = √169 = 13 m. 3.The midpoint has the mean of each coordinate: M = (0 + 122, 0 + 42, 0 + 32) = (6, 2, 1.5).
The midpoint has the mean of each coordinate: M = (6, 2, 1.5), halfway up the room. 4.From the sensor to the hook: SM = 6 − 02 − 41.5 − 3 = 6−2−1.5.
From the sensor to the hook: SM = 6−2−1.5. 5.(b) M is at (6, 2, 1.5), and SM = √62 + 22 + 1.52 = √36 + 4 + 2.25 = √42.25 = 6.5 m. Check: 6.52 = 42.25.
(b) SM = √36 + 4 + 2.25 = √42.25 = 6.5 m.
Answer: (a) 13 m; (b) M(6, 2, 1.5), and 6.5 m from the sensor
Common mistakes
- Using only the floor: √122 + 42 = √160 ≈ 12.6 m. That is the diagonal of the floor; the cable also climbs 3 m, so the height enters the sum as a third square.
- Putting the hook at a height of 3 m because the light is at the ceiling. The cable starts on the floor, so halfway along it is halfway up, at 1.5 m.