Mathematics Questions and Answers – Three Dimensional Geometry – Distance between Two Points

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Three Dimensional Geometry – Distance between Two Points”.

1. Do we have distance formula in 3-D geometry also?
a) True
b) False
View Answer

Answer: a
Explanation: Yes, we have distance formula in 3-D geometry also. Distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
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2. Find the distance between two points (5, 6, 7) and (2, 6, 3).
a) 3 units
b) 0 units
c) 4 units
d) 5 units
View Answer

Answer: d
Explanation: We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
So, distance between two points (5, 6, 7) and (2, 6, 3) will be \(\sqrt{(5-2)^2+(6-6)^2+(7-3)^2}\) = \(\sqrt{(3)^2+(4)^2}\) = 5 units.

3. The points A (3, 2, 1), B (5, 3, -2) and C (-1, 0, 7) are collinear.
a) True
b) False
View Answer

Answer: a
Explanation: We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
Distance AB = \(\sqrt{(3-5)^2+(2-3)^2+(1+2)^2} = \sqrt{14}\)
Distance BC = \(\sqrt{(5+1)^2+(3-0)^2+(-2-7)^2} = 3\sqrt{14}\)
Distance AC = \(\sqrt{(3+1)^2+(2-0)^2+(1-7)^2} = 2\sqrt{14}\)
Since AB+AC = BC so, the three points are collinear.
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4. The three points A (1, 2, 3), B (3, 1, 2), C (2, 3, 1) form ________________
a) equilateral triangle
b) right angled triangle
c) isosceles triangle
d) right angled isosceles triangle
View Answer

Answer: a
Explanation: We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
Distance AB = \(\sqrt{(1-3)^2+(2-1)^2+(3-2)^2} = \sqrt{6}\)
Distance BC = \(\sqrt{(3-2)^2+(1-3)^2+(2-1)^2} = \sqrt{6}\)
Distance CA = \(\sqrt{(2-1)^2+(3-2)^2+(1-3)^2} = \sqrt{6}\)
Since AB=BC=CA so, it forms equilateral triangle.

5. The three points A (3, 0, 3), B (5, 3, 2), C (6, 5, 5) form ________________
a) equilateral triangle
b) right angled triangle
c) isosceles triangle
d) right angled isosceles triangle
View Answer

Answer: c
Explanation: We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
Distance AB = \(\sqrt{(3-5)^2+(0-3)^2+(3-2)^2} = \sqrt{14}\)
Distance BC =\(\sqrt{(5-6)^2+(3-5)^2+(2-5)^2} = \sqrt{14}\)
Distance AC =\(\sqrt{(3-6)^2+(0-5)^2+(3-5)^2} = \sqrt{38}\)
Since AB =BC so, it forms isosceles triangle.
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6. The three points A (7, 0, 10), B (6, -1, 6), C (9, -4, 6) form ________________
a) equilateral triangle
b) right angled triangle
c) isosceles triangle
d) right angled isosceles triangle
View Answer

Answer: d
Explanation: We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
Distance AB = \(\sqrt{(7-6)^2+(0+1)^2+(10-6)^2} = \sqrt{18}\)
Distance BC = \(\sqrt{(6-9)^2+(-1+4)^2+(6-6)^2} = \sqrt{18}\)
Distance AC = \(\sqrt{(7-9)^2+(0+4)^2+(10-6)^2} = \sqrt{36}\)
Since AB = BC and AB2+BC2 = AC2 so, it forms right angled isosceles triangle.

7. The points A (1, 2, -1), B (5, -2, 1), C (8, -7, 4), D (4, -3, 2) form_____________
a) trapezium
b) rhombus
c) square
d) parallelogram
View Answer

Answer: d
Explanation: We know, distance between two points (x1, y1, z1) and (x2, y2, z2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
Distance AB = \(\sqrt{(1-5)^2+(2+2)^2+(-1-1)^2}\) = 6
Distance BC = \(\sqrt{(5-8)^2+(-2+7)^2+(1-4)^2} = \sqrt{43}\)
Distance CD = \(\sqrt{(8-4)^2+(-7+3)^2+(4-2)^2}\) = 6
Distance AD = \(\sqrt{(1-4)^2+(2+3)^2+(-1-2)^2} = \sqrt{43}\)
Since AB=CD and BC=AD i.e. opposite two sides are equal so, it is a parallelogram.
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter