3D Coordinates and Distance Formula

Introduction to 3D Coordinate System

Just as we use the ordered pair (x,y) to locate a point on a 2D plane, we use an ordered triplet (x, y, z) to locate a point in 3D space. This system is built upon three mutually perpendicular axes: the x-axis, y-axis, and z-axis, which intersect at the origin (0,0,0).

• Coordinate Planes: The axes define three coordinate planes that act as reference surfaces:

  • The xy-plane (where z=0)

  • The yz-plane (where x=0)

  • The xz-plane (where y=0)

• Octants: These three planes divide the space into eight regions called octants. The sign of the coordinates (x,y,z) determines the octant in which a point lies.

  

Coordinates of a Point in Space

A point P in space is represented by P(x, y, z). Here, x, y, and z are the perpendicular distances of P from the yz-plane, xz-plane, and xy-plane, respectively.

Distance Formula in 3D

The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is a direct extension of the 2D formula, derived using the Pythagorean theorem twice.

$$PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$

The distance of a point P(x, y, z) from the origin is:

$$OP = \sqrt{x^2 + y^2 + z^2}$$

Example 1: Basic Distance Formula

Question: Find the distance between the points P(1, -3, 4) and Q(-4, 1, 2).

Solution:

Using the distance formula $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:

$$PQ = \sqrt{(-4-1)^2 + (1-(-3))^2 + (2-4)^2}$$

$$= \sqrt{(-5)^2 + (4)^2 + (-2)^2}$$

$$= \sqrt{25 + 16 + 4} = \sqrt{45} = \mathbf{3\sqrt{5}}$$

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Example 2: Identifying the Octant

Question: In which octant does the point (-3, 1, -2) lie?

Solution:

We look at the signs of the coordinates (x, y, z), which are (-, +, -).

• x is negative

• y is positive

• z is negative

Referring to the octant sign convention table, a (-, +, -) configuration corresponds to the VI octant.

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Example 3: Collinearity of Points

Question: Show that the points P(-2, 3, 5), Q(1, 2, 3) and R(7, 0, -1) are collinear.

Solution:

Three points are collinear if the sum of the distances between two pairs of points equals the distance between the third pair. We calculate all three distances:

$PQ = \sqrt{(1-(-2))^2 + (2-3)^2 + (3-5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9+1+4} = \mathbf{\sqrt{14}}$.

$QR = \sqrt{(7-1)^2 + (0-2)^2 + (-1-3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36+4+16} = \sqrt{56} = \mathbf{2\sqrt{14}}$.

$PR = \sqrt{(7-(-2))^2 + (0-3)^2 + (-1-5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81+9+36} = \sqrt{126} = \mathbf{3\sqrt{14}}$.

Now, we check the condition: $PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14}$, which is equal to PR. Since $PQ+QR=PR$, the points are collinear. ✅

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Example 4: Identifying an Isosceles Triangle

Question: Show that the points A(0,7,10), B(-1,6,6) and C(-4,9,6) form an isosceles triangle.

Solution:

We find the lengths of the three sides:

$AB = \sqrt{(-1-0)^2+(6-7)^2+(6-10)^2} = \sqrt{1+1+16} = \sqrt{18} = \mathbf{3\sqrt{2}}$.

$BC = \sqrt{(-4-(-1))^2+(9-6)^2+(6-6)^2} = \sqrt{(-3)^2+3^2+0^2} = \sqrt{9+9+0} = \sqrt{18} = \mathbf{3\sqrt{2}}$.

$AC = \sqrt{(-4-0)^2+(9-7)^2+(6-10)^2} = \sqrt{16+4+16} = \sqrt{36} = \mathbf{6}$.

Since two sides have equal length ($AB=BC$), the triangle is isosceles.

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Example 5: Identifying a Right-Angled Triangle

Question: Show that the points from the previous example, A(0,7,10), B(-1,6,6) and C(-4,9,6), also form a right-angled triangle.

Solution:

We check if the side lengths satisfy the converse of the Pythagorean theorem, $a^2+b^2=c^2$. From the previous example, we have the side lengths squared:

• $AB^2 = 18$

• $BC^2 = 18$

• $AC^2 = 36$

We check the sum of the squares of the two shorter sides: $AB^2 + BC^2 = 18+18=\mathbf{36}$.

This sum is equal to the square of the longest side, $AC^2$. Since $AB^2 + BC^2 = AC^2$, the triangle is right-angled, with the right angle at vertex B.

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Example 6: Locus of Equidistant Points

Question: Find the equation of the set of points P which are equidistant from the points A(1,2,3) and B(3,2,-1).

Solution:

Let the point P be $(x,y,z)$. The condition is $PA=PB$, which is equivalent to $PA^2=PB^2$. This avoids square roots.

$$(x-1)^2+(y-2)^2+(z-3)^2 = (x-3)^2+(y-2)^2+(z-(-1))^2$$

Notice that the $(y-2)^2$ term is on both sides and cancels out:

$$(x-1)^2+(z-3)^2 = (x-3)^2+(z+1)^2$$

Expand the terms:

$$x^2-2x+1+z^2-6z+9 = x^2-6x+9+z^2+2z+1$$

Cancel $x^2, z^2, 9,$ and $1$ from both sides:

$$-2x-6z = -6x+2z$$

Bring all terms to one side:

$$4x-8z=0 \implies \mathbf{x-2z=0}$$

This is the equation of a plane that perpendicularly bisects the segment AB.

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Example 7: Finding a Point on an Axis

Question: Find the coordinates of the point on the y-axis which is at a distance of $5\sqrt{2}$ from the point P(3, -2, 5).

Solution:

Step 1: Define the point on the y-axis. Any point on the y-axis has its x and z coordinates equal to zero. So, let the point be A(0, y, 0).

Step 2: Use the distance formula. We are given the distance $AP = 5\sqrt{2}$, so $AP^2 = (5\sqrt{2})^2 = 50$.

$(3-0)^2 + (-2-y)^2 + (5-0)^2 = 50$

$9 + (y+2)^2 + 25 = 50$

$34 + (y+2)^2 = 50$

Step 3: Solve for y.

$(y+2)^2 = 16$

$y+2 = \pm \sqrt{16} = \pm 4$.

This gives two possible values for y:

• $y+2 = 4 \implies y=2$

• $y+2 = -4 \implies y=-6$

The points are (0, 2, 0) and (0, -6, 0).

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Example 8: Identifying a Parallelogram

Question: Show that the points A(-1,2,1), B(1,-2,5), C(4,-7,8) and D(2,-3,4) are the vertices of a parallelogram.

Solution:

A key property of a parallelogram is that its diagonals bisect each other. This means they share the same midpoint.

Step 1: Find the midpoint of the diagonal AC.

Midpoint AC = $(\frac{-1+4}{2}, \frac{2+(-7)}{2}, \frac{1+8}{2}) = \mathbf{(3/2, -5/2, 9/2)}$.

Step 2: Find the midpoint of the diagonal BD.

Midpoint BD = $(\frac{1+2}{2}, \frac{-2+(-3)}{2}, \frac{5+4}{2}) = \mathbf{(3/2, -5/2, 9/2)}$.

Since the midpoints of both diagonals are identical, the diagonals bisect each other, and therefore the vertices form a parallelogram.