Three-Dimensional Geometry — Chapter Test | Class 12 IMO

Q1

A line makes angle θ with the x-axis, and angle φ with the y-axis. What angle does it make with the z-axis if θ = φ = 60°?

cos²60° + cos²60° + cos²γ = 1. 1/4 + 1/4 + cos²γ = 1 → 1/2 + cos²γ = 1 → cos²γ = 1/2. cos γ = ±1/√2. So γ = 45° or 135°.

Q2

Which of the following represents the x-axis in 3D geometry?

The x-axis passes through origin (0,0,0) and has direction ratios 1, 0, 0. Hence, x/1 = y/0 = z/0.

Q3

A drone at a Diwali festival flies along the vector r = 2i + j + k + μ(i + j). What is its shortest distance from a static firework rocket launched vertically along the z-axis (r = λk)?

Line 1: a₁ = 2i + j + k, b₁ = i + j. Line 2: a₂ = 0, b₂ = k. Shortest distance = |(a₂ − a₁) · (b₁ × b₂)| / |b₁ × b₂|. a₂ − a₁ = −2i − j − k. b₁ × b₂ = (i+j) × k = −j + i = i − j. (a₂ − a₁) · (b₁ × b₂) = (−2)(1) + (−1)(−1) + 0 = −2 + 1 = −1. |b₁ × b₂| = √2. SD = |−1| / √2 = 1/√2.

Q4

The direction cosines of a line are proportional to 2, 3, 6. The direction cosine along z-axis is:

Magnitude = √(2²+3²+6²)=7. Hence direction cosines are 2/7, 3/7, 6/7. So z-direction cosine is 6/7.

Q5

What is the angle between the planes 3x + 4y − 5z = 9 and 2x + 6y + 6z = 7?

Normals: n₁ = (3, 4, −5), n₂ = (2, 6, 6). Dot product = 3(2) + 4(6) + (−5)(6) = 6 + 24 − 30 = 0. Planes are perpendicular (90° or cos⁻¹(0)).

Q6

The normal vector to the plane 2x−3y+6z−5=0 is:

Coefficients of x, y, z give the normal vector.

Q7

Sachin is designing a roof structure for a community center in New Delhi. The roof plane passes through three pillars located at (2, 2, −1), (3, 4, 2), and (7, 0, 6) in the site grid. What is the Cartesian equation of the roof plane?

Points A(2, 2, −1), B(3, 4, 2), C(7, 0, 6). Vectors AB = (1, 2, 3), AC = (5, −2, 7). Normal n = AB × AC = i(14 − (−6)) − j(7 − 15) + k(−2 − 10) = 20i + 8j − 12k. Simplifying DRs: 5, 2, −3. Equation: 5(x−2) + 2(y−2) − 3(z+1) = 0 → 5x − 10 + 2y − 4 − 3z − 3 = 0 → 5x + 2y − 3z = 17.

Q8

Two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are parallel if:

For planes to be parallel, their normal vectors must be proportional, meaning their direction ratios are in constant proportion: a₁/a₂ = b₁/b₂ = c₁/c₂.

Q9

The coordinates of the point on the line (x−1)/2 = (y+1)/−3 = z/1 which is at a distance of √14 from the point (1, −1, 0) are:

General point is (2λ+1, −3λ−1, λ). Distance squared from (1, −1, 0) is (2λ)² + (−3λ)² + λ² = 14λ². Given distance is √14, so 14λ² = 14 → λ² = 1 → λ = ±1. If λ = 1, point is (3, −4, 1). If λ = −1, point is (−1, 2, −1).

Q10

Find the equation of the line passing through (1, 2, 3) and perpendicular to the planes x − y + z = 1 and 2x + y − 3z = 4.

Direction of line is cross product of normal to the planes: n₁ = (1, −1, 1), n₂ = (2, 1, −3). n₁ × n₂ = i(3−1) − j(−3−2) + k(1+2) = 2i + 5j + 3k. DRs are 2, 5, 3. Equation is (x−1)/2 = (y−2)/5 = (z−3)/3.

Q11

The sum of the squares of direction cosines of a straight line is always equal to:

By definition, the sum of squares of direction cosines is l² + m² + n² = 1.

Q12

The planes 2x + 3y + 4z = 5 and 4x + 6y + 8z = 12 are:

Ratio of normals: 2/4 = 3/6 = 4/8 = 1/2. Since the normals are proportional, the planes are parallel.

Q13

The equation of plane with intercepts 2, 4 and 8 on axes is:

Directly from intercept form.

Q14

A mobile tower is at origin (0,0,0). An antenna aims exactly along the vector 2i + j + 2k. A signal receiver is placed at (4, 5, 6). What is the shortest distance from the receiver to the line of the antenna signal?

Line: r = λ(2i + j + 2k). Point P = (4, 5, 6). OP = 4i + 5j + 6k. b = 2i + j + 2k. Distance = |OP × b| / |b|. OP × b = i(10−6) − j(8−12) + k(4−10) = 4i + 4j − 6k. |OP × b| = √(16+16+36) = √68. |b| = 3. Distance = √(68)/3. i(5*2 - 6*1) = 4. -j(4*2 - 6*2) = -j(8-12) = 4j. k(4*1 - 5*2) = k(4-10) = -6k. Correct. √68 / 3. We adjust an option: √(68)/3. We will write the nearest correct or correct the option. We will correct the option '√(65)/3' to '√(68)/3'. I must output standard. We pick '√(68)/3' as option 1. We'll modify the text here.

Q15

Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.

Image formula: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c = −2(ax₁+by₁+cz₁−d) / (a²+b²+c²). x₁=1, y₁=2, z₁=−1. a=3, b=−5, c=4, d=5. Val = 3(1) − 5(2) + 4(−1) − 5 = 3 − 10 − 4 − 5 = −16. a²+b²+c² = 9 + 25 + 16 = 50. Ratio = −2(−16)/50 = 32/50 = 16/25. x = 1 + 3(16/25) = 73/25. y = 2 − 5(16/25) = 2 − 16/5 = −6/5. z = −1 + 4(16/25) = 39/25.