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Introduction to Three Dimensional Geometry - Distance between Two Points

Grade 11CBSE

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

🔑Concepts

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The 3D Coordinate System is defined by three mutually perpendicular lines passing through a common point called the origin O(0,0,0)O(0,0,0). These lines are the X, Y, and Z axes. Visually, these axes divide the entire space into eight distinct regions called octants, similar to how 2D axes divide a plane into four quadrants.

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A point PP in three-dimensional space is represented as an ordered triplet (x,y,z)(x, y, z). Geometrically, xx is the perpendicular distance from the YZYZ-plane, yy is the distance from the XZXZ-plane, and zz is the distance from the XYXY-plane. If you imagine a box with one corner at the origin and the opposite corner at PP, the edges of the box represent these coordinates.

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The Distance Formula in 3D is an extension of the Pythagorean theorem. To find the distance between P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2), we calculate the square root of the sum of the squares of the differences of their corresponding coordinates. Visually, this distance represents the length of the straight line segment (or the space diagonal of a rectangular prism) connecting the two points.

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The distance of any point P(x,y,z)P(x, y, z) from the origin O(0,0,0)O(0, 0, 0) is simplified to x2+y2+z2\sqrt{x^2 + y^2 + z^2}, as the starting coordinates are all zero.

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Collinearity of Points can be tested using the distance formula. Three points A,B,A, B, 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, such as AB+BC=ACAB + BC = AC. Visually, this means the points don't form a triangle, but a single line segment.

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Geometric Properties of figures can be verified using distances. For instance, four points form a parallelogram if the opposite sides are equal (AB=CDAB=CD and BC=DABC=DA), and they form a rhombus if all four sides are equal (AB=BC=CD=DAAB=BC=CD=DA). A right-angled triangle is verified if the square of the longest side equals the sum of the squares of the other two sides.

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The position of a point relative to coordinate planes is important for visualization: points on the XYXY-plane always have z=0z = 0, points on the YZYZ-plane have x=0x = 0, and points on the ZXZX-plane have y=0y = 0.

📐Formulae

Distance between 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 the origin O(0,0,0)O(0, 0, 0): OP=x2+y2+z2OP = \sqrt{x^2 + y^2 + z^2}

Condition for collinearity of A,B,CA, B, C: AB+BC=ACAB + BC = AC (or any permutation where the sum of two segments equals the third)

Condition for Right-angled triangle: AB2+BC2=AC2AB^2 + BC^2 = AC^2 (Pythagoras Theorem)

💡Examples

Problem 1:

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

Solution:

Step 1: Identify the 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).\nStep 2: Apply the 3D distance formula: PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}.\nStep 3: Substitute the values: PQ=(−4−1)2+(1−(−3))2+(2−4)2PQ = \sqrt{(-4 - 1)^2 + (1 - (-3))^2 + (2 - 4)^2}.\nStep 4: Simplify: PQ=(−5)2+(4)2+(−2)2=25+16+4=45PQ = \sqrt{(-5)^2 + (4)^2 + (-2)^2} = \sqrt{25 + 16 + 4} = \sqrt{45}.\nStep 5: Final simplification: PQ=35PQ = 3\sqrt{5} units.

Explanation:

The distance is found by substituting the coordinates into the 3D distance formula, which calculates the length of the vector connecting the two points in space.

Problem 2:

Show 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 collinear.

Solution:

Step 1: Calculate distance AB=(1−(−2))2+(2−3)2+(3−5)2=32+(−1)2+(−2)2=9+1+4=14AB = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}.\nStep 2: Calculate distance BC=(7−1)2+(0−2)2+(−1−3)2=62+(−2)2+(−4)2=36+4+16=56=214BC = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}.\nStep 3: Calculate distance AC=(7−(−2))2+(0−3)2+(−1−5)2=92+(−3)2+(−6)2=81+9+36=126=314AC = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}.\nStep 4: Check if AB+BC=ACAB + BC = AC. We have 14+214=314\sqrt{14} + 2\sqrt{14} = 3\sqrt{14}, which matches the value of ACAC.

Explanation:

To prove collinearity, we calculate the distances between all three pairs of points. If the sum of the two shorter distances equals the longest distance, the points lie on a single straight line.