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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 formed by three mutually perpendicular axes: XX, YY, and ZZ. Any point PP in space is uniquely identified by an ordered triplet (x,y,z)(x, y, z), which represents the signed distances from the YZYZ, ZXZX, and XYXY planes respectively.

3D coordinate axes X, Y, Z showing a point P in space.
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The distance formula in 3D is an extension of the 2D Pythagorean distance. For points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2), the distance PQPQ is the square root of the sum of the squares of the differences of their corresponding coordinates.

A cuboid illustrating the space diagonal PQ representing distance in 3D.
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Geometric properties such as isosceles, equilateral, or right-angled triangles can be verified by calculating the lengths of the sides using the distance formula. For instance, a triangle is isosceles if at least two side lengths are equal.

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A point P(x,y,z)P(x, y, z) lies on the XYXY-plane if its zz-coordinate is zero (z=0z=0), on the YZYZ-plane if x=0x=0, and on the ZXZX-plane if 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.

Problem 3:

Verify if the points A(0,7,10)A(0, 7, 10), B(−1,6,6)B(-1, 6, 6) and C(−4,9,6)C(-4, 9, 6) form an isosceles right-angled triangle.

Right-angled isosceles triangle ABC with side lengths 3√2.

Solution:

  1. Calculate side ABAB: AB=(−1−0)2+(6−7)2+(6−10)2=1+1+16=18=32AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2} units.

  2. Calculate side BCBC: BC=(−4−(−1))2+(9−6)2+(6−6)2=9+9+0=18=32BC = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2} units.

  3. Calculate side ACAC: AC=(−4−0)2+(9−7)2+(6−10)2=16+4+16=36=6AC = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 units.

Since AB=BC=18AB = BC = \sqrt{18}, the triangle is isosceles. Also, AB2+BC2=18+18=36AB^2 + BC^2 = 18 + 18 = 36 and AC2=36AC^2 = 36. Since AB2+BC2=AC2AB^2 + BC^2 = AC^2, the triangle is right-angled at BB.

Explanation:

We use the distance formula to find all three side lengths. Comparing the lengths reveals that two sides are equal (isosceles) and the square of the longest side equals the sum of squares of the other two (Pythagorean theorem).

Problem 4:

Find the equation of the set of points P(x,y,z)P(x, y, z) such that its distance from the point A(3,4,−5)A(3, 4, -5) is equal to its distance from the point B(−2,1,4)B(-2, 1, 4).

Points A and B separated by a plane representing the locus of point P.

Solution:

Let P(x,y,z)P(x, y, z) be the point. Given PA=PBPA = PB. So, PA2=PB2PA^2 = PB^2. (x−3)2+(y−4)2+(z+5)2=(x+2)2+(y−1)2+(z−4)2(x-3)^2 + (y-4)^2 + (z+5)^2 = (x+2)^2 + (y-1)^2 + (z-4)^2 Expanding the squares: (x2−6x+9)+(y2−8y+16)+(z2+10z+25)=(x2+4x+4)+(y2−2y+1)+(z2−8z+16)(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 10z + 25) = (x^2 + 4x + 4) + (y^2 - 2y + 1) + (z^2 - 8z + 16) Cancelling x2,y2,z2x^2, y^2, z^2 from both sides: −6x−8y+10z+50=4x−2y−8z+21-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21 Rearranging terms: 10x+6y−18z−29=010x + 6y - 18z - 29 = 0

Explanation:

This problem describes the locus of points equidistant from two fixed points, which is the perpendicular bisector plane of the segment AB. We equate the squares of the distances to remove the square roots.