IMO Practice Test: Introduction to Three Dimensional Geometry | Class 11 Mathematics

Question 1 of 6 hard

If the origin is the centroid of the triangle with vertices (2a, 2, 6), (−4, 3b, −10) and (8, 14, 2c), find the values of a, b, and c.

−2, 16/3, −2

−2, −16/3, 2

2, 16/3, 2

2, −16/3, −2

Explanation: Centroid is (0,0,0). For x: (2a−4+8)/3 = 0 → 2a = −4 → a = −2. For y: (2+3b+14)/3 = 0 → 3b = −16 → b = −16/3. For z: (6−10+2c)/3 = 0 → 2c = 4 → c = 2.

Question 2 of 6 hard

What is the primary locus of a point which moves such that its absolute distance from the x-axis is always equal to its absolute distance from the y-axis?

x² = y²

y² = z²

x² + y² = z²

x² + z² = y² + z²

Explanation: Distance from the x-axis is √(y² + z²). Distance from the y-axis is √(x² + z²). Equating them: √(y² + z²) = √(x² + z²). Squaring both sides gives y² + z² = x² + z² → x² = y².

Question 3 of 6 hard

Define accurately the mathematical locus formed by an unconstrained point perpetually balancing a uniform spatial distance of 5 defined units radially from absolute origin.

x² + y² = 5

x² + y² + z² = 25

x² + y² + z² = 5

x² + z² = 25

Explanation: Formulating distance strictly implies √(x² + y² + z²) = 5 algebraically. Squaring absolutely everything logically clears the root framing the spatial sphere accurately as x² + y² + z² = 25.

Question 4 of 6 hard

For points A(1,2,3), B(4,5,6), midpoint M and origin O, OM² equals:

151/4

155/4

159/4

165/4

Explanation: Midpoint M = (5/2, 7/2, 9/2). Then OM² = (5/2)² + (7/2)² + (9/2)² = (25 + 49 + 81)/4 = 155/4.

Question 5 of 6 hard

A cubical room corner is represented by origin and opposite corner by (4,4,4). The body diagonal length is:

4√3

8

12

16

Explanation: Distance = √(4²+4²+4²)=√48=4√3.

Question 6 of 6 hard

Find the z-coordinate of the point on the z-axis which is equidistant from the points (1, 5, 7) and (5, 1, −4).

3/2

2

5/2

−3/2

Explanation: Let point be P(0, 0, z). P to (1, 5, 7): 1² + 5² + (z−7)² = z² − 14z + 75. P to (5, 1, −4): 5² + 1² + (z+4)² = z² + 8z + 42. Equating: z² − 14z + 75 = z² + 8z + 42 → 22z = 33 → z = 33/22 = 3/2.

IMO Practice Test — Introduction to Three Dimensional Geometry