# 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

Explanation: Yes, we have distance formula in 3-D geometry also. Distance between two points (x1, y1, z1) and (x2, y2, z2) is $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$.

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

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

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.

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

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

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.

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

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

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. 