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

Explanation: Yes, we have distance formula in 3-D geometry also. Distance between two points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) 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

View Answer

Explanation: We know, distance between two points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) 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

Explanation: We know, distance between two points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) 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

View Answer

Explanation: We know, distance between two points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) 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

Explanation: We know, distance between two points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) 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

View Answer

Explanation: We know, distance between two points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) 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 AB

^{2}+BC

^{2}= AC

^{2}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

Explanation: We know, distance between two points (x

_{1}, y

_{1}, z

_{1}) and (x

_{2}, y

_{2}, z

_{2}) 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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