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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

The 3D Cartesian Coordinate System: Space is defined by three mutually perpendicular lines called the xx, yy, and zz axes that meet at a point called the origin O(0,0,0)O(0, 0, 0). Visually, this is similar to the corner of a room where the floor meet two walls; the lines of intersection represent the axes, and any point in the room can be located relative to that corner.

Coordinate Planes and Octants: The three axes form three coordinate planes: the XYXY-plane (horizontal), the YZYZ-plane (vertical side), and the ZXZX-plane (vertical front). These planes divide the space into eight regions called octants. Visually, think of a 2D grid being split by a horizontal sheet, resulting in four sections above the sheet and four sections below.

Coordinates of a Point: A point PP in 3D space is assigned an ordered triple (x,y,z)(x, y, z). The xx-coordinate is the perpendicular distance from the YZYZ-plane, yy is the distance from the ZXZX-plane, and zz is the distance from the XYXY-plane. Visually, to reach a point (2,3,4)(2, 3, 4), you move 2 units along the xx-axis, 3 units parallel to the yy-axis, and 4 units up parallel to the zz-axis.

The 3D Distance Formula: 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 PQPQ. It is an extension of the 2D Pythagorean theorem. Visually, the distance PQPQ represents the length of the longest diagonal of a rectangular box (cuboid) whose edges are parallel to the coordinate axes and have lengths x2x1|x_2 - x_1|, y2y1|y_2 - y_1|, and z2z1|z_2 - z_1|.

Distance from the Origin: A specific application of the distance formula occurs when one point is the origin O(0,0,0)O(0, 0, 0). The distance of any point P(x,y,z)P(x, y, z) from the origin is simply the square root of the sum of the squares of its coordinates. Visually, this is the straight-line distance from the very center of the coordinate system to the point in space.

Collinearity in 3D: Three points AA, BB, and CC are collinear (lie on the same straight line) if the sum of the lengths of any two segments equals the length of the third segment. For example, if AB+BC=ACAB + BC = AC, the points are collinear. This is verified by calculating three separate distances using the distance formula.

Properties of Geometric Figures: The distance formula is used to classify triangles and quadrilaterals in 3D space. For instance, if all three side lengths calculated are equal, the triangle is equilateral; if the sum of the squares of two sides equals the square of the third, it is a right-angled triangle.

📐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=(x2x1)2+(y2y1)2+(z2z1)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=(41)2+(1(3))2+(24)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=(10)2+(67)2+(610)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+(96)2+(66)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+(79)2+(106)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.