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Introduction to Three-Dimensional Geometry - Distance between Two Points

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The 3D Coordinate System consists of three mutually perpendicular axes: XX, YY, and ZZ, which meet at the origin O(0,0,0)O(0, 0, 0). Any point PP in space is represented by the coordinates (x,y,z)(x, y, z). These coordinates represent the signed distances of the point from the YZYZ, ZXZX, and XYXY planes respectively.

3D coordinate axes showing a point P with respect to the origin.
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The distance between two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2) is the length of the line segment joining them, calculated using the 3D extension of the Pythagorean theorem: (x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}.

A line segment PQ in 3D space representing the distance d.
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Distance from Coordinate Planes: The distance of point P(x,y,z)P(x, y, z) from the XYXY-plane is ∣z∣|z|, from the YZYZ-plane is ∣x∣|x|, and from the ZXZX-plane is ∣y∣|y|.

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Distance from Coordinate Axes: The distance of point P(x,y,z)P(x, y, z) from the XX-axis is y2+z2\sqrt{y^2 + z^2}, from the YY-axis is x2+z2\sqrt{x^2 + z^2}, and from the ZZ-axis is x2+y2\sqrt{x^2 + y^2}.

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Collinearity of Points: Three points AA, BB, and CC are collinear if the sum of the lengths of any two segments equals the length of the third segment (e.g., AB+BC=ACAB + BC = AC).

📐Formulae

Distance between points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2): d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Distance of point P(x,y,z)P(x, y, z) from origin O(0,0,0)O(0, 0, 0): OP=x2+y2+z2OP = \sqrt{x^2 + y^2 + z^2}

Distance of point P(x,y,z)P(x, y, z) from the XX-axis: dx=y2+z2d_x = \sqrt{y^2 + z^2}

Distance of point P(x,y,z)P(x, y, z) from the YY-axis: dy=x2+z2d_y = \sqrt{x^2 + z^2}

Distance of point P(x,y,z)P(x, y, z) from the ZZ-axis: dz=x2+y2d_z = \sqrt{x^2 + y^2}

💡Examples

Problem 1:

Find the distance between the points A(1,−3,4)A(1, -3, 4) and B(−4,1,2)B(-4, 1, 2).

Solution:

  1. Identify coordinates: (x1,y1,z1)=(1,−3,4)(x_1, y_1, z_1) = (1, -3, 4) and (x2,y2,z2)=(−4,1,2)(x_2, y_2, z_2) = (-4, 1, 2).
  2. Substitute into the formula: AB=(−4−1)2+(1−(−3))2+(2−4)2AB = \sqrt{(-4 - 1)^2 + (1 - (-3))^2 + (2 - 4)^2}.
  3. Simplify the terms: AB=(−5)2+(4)2+(−2)2AB = \sqrt{(-5)^2 + (4)^2 + (-2)^2}.
  4. Calculate squares: AB=25+16+4AB = \sqrt{25 + 16 + 4}.
  5. Final result: AB=45=35AB = \sqrt{45} = 3\sqrt{5} units.

Explanation:

We use the standard 3D distance formula by calculating the difference between corresponding xx, yy, and zz coordinates, squaring them, adding them together, and taking the square root.

Problem 2:

Show that the points P(0,7,10)P(0, 7, 10), Q(−1,6,6)Q(-1, 6, 6), and R(−4,9,6)R(-4, 9, 6) form an isosceles right-angled triangle.

Solution:

  1. Calculate PQPQ: PQ=(−1−0)2+(6−7)2+(6−10)2=(−1)2+(−1)2+(−4)2=1+1+16=18PQ = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1+1+16} = \sqrt{18}.
  2. Calculate QRQR: QR=(−4−(−1))2+(9−6)2+(6−6)2=(−3)2+32+02=9+9+0=18QR = \sqrt{(-4-(-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9+9+0} = \sqrt{18}.
  3. Calculate RPRP: RP=(0−(−4))2+(7−9)2+(10−6)2=42+(−2)2+42=16+4+16=36=6RP = \sqrt{(0-(-4))^2 + (7-9)^2 + (10-6)^2} = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16+4+16} = \sqrt{36} = 6.
  4. Check Isosceles property: Since PQ=QR=18PQ = QR = \sqrt{18}, the triangle is isosceles.
  5. Check Right-angled property (Pythagoras): PQ2+QR2=(18)2+(18)2=18+18=36PQ^2 + QR^2 = (\sqrt{18})^2 + (\sqrt{18})^2 = 18 + 18 = 36. Since RP2=62=36RP^2 = 6^2 = 36, PQ2+QR2=RP2PQ^2 + QR^2 = RP^2.

Explanation:

To verify the type of triangle, we find the lengths of all three sides. Matching lengths indicate an isosceles triangle, and satisfying the Pythagorean theorem confirms it is right-angled.

Problem 3:

Verify using the distance formula that the points A(2,3,5)A(2, 3, 5), B(1,2,3)B(1, 2, 3), and C(7,0,−1)C(7, 0, -1) are the vertices of a right-angled triangle. Also find the length of the hypotenuse.

Triangle ABC plotted in a 3D coordinate context.

Solution:

  1. Calculate AB2AB^2: (1−2)2+(2−3)2+(3−5)2=(−1)2+(−1)2+(−2)2=1+1+4=6(1-2)^2 + (2-3)^2 + (3-5)^2 = (-1)^2 + (-1)^2 + (-2)^2 = 1 + 1 + 4 = 6
  2. Calculate BC2BC^2: (7−1)2+(0−2)2+(−1−3)2=62+(−2)2+(−4)2=36+4+16=56(7-1)^2 + (0-2)^2 + (-1-3)^2 = 6^2 + (-2)^2 + (-4)^2 = 36 + 4 + 16 = 56
  3. Calculate AC2AC^2: (7−2)2+(0−3)2+(−1−5)2=52+(−3)2+(−6)2=25+9+36=70(7-2)^2 + (0-3)^2 + (-1-5)^2 = 5^2 + (-3)^2 + (-6)^2 = 25 + 9 + 36 = 70 Since AB2+BC2=6+56=62≠70AB^2 + BC^2 = 6 + 56 = 62 \neq 70, the initial verification shows it is not right-angled at BB. Re-checking the values: AC2=70AC^2 = 70, AB2=6AB^2 = 6, BC2=56BC^2 = 56. 6+56=626 + 56 = 62. Let's check if the points were meant to be different. Given these coordinates, the triangle is not right-angled as the sum of squares of two sides does not equal the third. The longest side is AC=70AC = \sqrt{70}.

Explanation:

To check for a right-angled triangle, we calculate the squares of the lengths of all three sides using the distance formula and apply the converse of the Pythagorean theorem (a2+b2=c2a^2 + b^2 = c^2).

Problem 4:

Find the point on the ZZ-axis which is equidistant from the points A(1,5,7)A(1, 5, 7) and B(5,1,−4)B(5, 1, -4).

Diagram showing point P on the Z-axis equidistant from points A and B.

Solution:

  1. Let the required point on the ZZ-axis be P(0,0,z)P(0, 0, z).
  2. Given PA=PBPA = PB, therefore PA2=PB2PA^2 = PB^2.
  3. PA2=(0−1)2+(0−5)2+(z−7)2=1+25+(z−7)2=26+z2−14z+49=z2−14z+75PA^2 = (0-1)^2 + (0-5)^2 + (z-7)^2 = 1 + 25 + (z-7)^2 = 26 + z^2 - 14z + 49 = z^2 - 14z + 75.
  4. PB2=(0−5)2+(0−1)2+(z−(−4))2=25+1+(z+4)2=26+z2+8z+16=z2+8z+42PB^2 = (0-5)^2 + (0-1)^2 + (z-(-4))^2 = 25 + 1 + (z+4)^2 = 26 + z^2 + 8z + 16 = z^2 + 8z + 42.
  5. Equating PA2PA^2 and PB2PB^2: z2−14z+75=z2+8z+42z^2 - 14z + 75 = z^2 + 8z + 42
  6. −14z−8z=42−75  ⟹  −22z=−33  ⟹  z=3322=32-14z - 8z = 42 - 75 \implies -22z = -33 \implies z = \frac{33}{22} = \frac{3}{2}. The point is P(0,0,1.5)P(0, 0, 1.5).

Explanation:

Any point on the ZZ-axis has xx and yy coordinates equal to 0. We use the distance formula to set the distance from PP to AA equal to the distance from PP to BB.