Three Dimensional Geometry — Class 12 Maths practice questions

Question 1

A drone delivery system operates in a warehouse modelled on a 3D coordinate system. The drone starts at depot A(1, −1, 2) and flies in a straight path with direction ratios 2 : 3 : −1. A sensor tower is located at point P(5, 5, 0).

(i) Write the vector equation of the flight path of the drone. [1]
(ii) Write the cartesian equation of the flight path of the drone. [1]
(iii) Find the foot of the perpendicular drawn from the sensor tower P(5, 5, 0) to the flight path of the drone. Hence find the shortest distance from the sensor tower to the flight path. [2]

A drone can only fly perpendicular to both conveyor belts. The warehouse manager needs to know:
(i) Whether the two conveyor belt lines are skew (non-intersecting, non-parallel). Verify by computing b⃗₁ × b⃗₂ and (B⃗ − A⃗) · (b⃗₁ × b⃗₂). [1 mark]
(ii) Find the vector b⃗₁ × b⃗₂ (the common perpendicular direction). [1 mark]
(iii) Find the shortest distance between the two conveyor belt lines, which is the minimum distance the drone must travel perpendicular to both belts. [2 marks]

Accepted Answer

(i) Verifying the lines are skew:

a⃗₁ = î − ĵ + 2k̂ (position vector of A), a⃗₂ = 2î + ĵ − k̂ (position vector of B)

(a⃗₂ − a⃗₁) = (2−1)î + (1−(−1))ĵ + (−1−2)k̂ = î + 2ĵ − 3k̂

First compute b⃗₁ × b⃗₂:

b⃗₁ × b⃗₂ = |î  ĵ  k̂ |
              |1  2  −1|
              |2  −1  3|

= î[(2)(3)−(−1)(−1)] − ĵ[(1)(3)−(−1)(2)] + k̂[(1)(−1)−(2)(2)]

= î[6 − 1] − ĵ[3 + 2] + k̂[−1 − 4]

= 5î − 5ĵ − 5k̂

Now check (a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂):

(î + 2ĵ − 3k̂) · (5î − 5ĵ − 5k̂)
= (1)(5) + (2)(−5) + (−3)(−5)
= 5 − 10 + 15 = 10 ≠ 0

∴ (a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂) = 10 ≠ 0, which confirms the two conveyor belt lines are skew (they neither intersect nor are parallel).

(ii) The common perpendicular direction is:

b⃗₁ × b⃗₂ = 5î − 5ĵ − 5k̂

∴ b⃗₁ × b⃗₂ = 5î − 5ĵ − 5k̂

(iii) Shortest distance between the two skew lines:

Using the formula: S.D. = |(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂)| / |b⃗₁ × b⃗₂|

Numerator = |(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂)| = |10| = 10

|b⃗₁ × b⃗₂| = |5î − 5ĵ − 5k̂| = √(5² + (−5)² + (−5)²) = √(25 + 25 + 25) = √75 = 5√3

∴ S.D. = 10 / (5√3) = 2/√3 = (2√3)/3

∴ Shortest distance between the two conveyor belt lines = (2√3)/3 units.

This is the minimum distance the drone must travel perpendicular to both belts.

Question 2

A drone delivery company uses a straight flight corridor modelled as a line in 3D space. The flight path passes through the depot at point A(1, 2, −1) and a waypoint at B(3, 6, 3). A second corridor passes through point C(2, 3, 1) with direction ratios 1 : 2 : 2.

Based on the above information, answer the following questions:

(i) Write the direction ratios of the first flight corridor (line AB). [1]

(ii) Write the vector equation of the second corridor (line through C with direction ratios 1 : 2 : 2). [1]

(iii) Determine whether the two flight corridors are parallel to each other. If not parallel, find the angle between them. [2]

OR

(iii) Find the cartesian equation of the first flight corridor (line AB) and verify that the point P(5, 10, 7) lies on it. [2]

Accepted Answer

(i) Direction ratios of line AB:

Direction ratios of line AB are proportional to (B − A), i.e.,
(3 − 1) : (6 − 2) : (3 − (−1)) = 2 : 4 : 4 = 1 : 2 : 2.

∴ Direction ratios of the first flight corridor are 1 : 2 : 2.

(ii) Vector equation of the second corridor:

The second corridor passes through C(2, 3, 1) with direction ratios 1 : 2 : 2, so its direction vector is b⃗₂ = î + 2ĵ + 2k̂ and position vector a⃗₂ = 2î + 3ĵ + k̂.

∴ Vector equation of the second corridor is
r⃗ = (2î + 3ĵ + k̂) + μ(î + 2ĵ + 2k̂), where μ ∈ ℝ.

(iii) [Main option] Checking whether the corridors are parallel; if not, finding the angle between them:

Direction vector of first corridor: b⃗₁ = î + 2ĵ + 2k̂ (using reduced DRs from part (i)).
Direction vector of second corridor: b⃗₂ = î + 2ĵ + 2k̂.

For two lines to be parallel, b⃗₁ = λb⃗₂ for some scalar λ.

Here b⃗₁ = 1·b⃗₂ (with λ = 1), so b⃗₁ and b⃗₂ are parallel vectors.

∵ b⃗₁ ∥ b⃗₂, the two flight corridors are parallel to each other.

∴ The two flight corridors are parallel; the angle between them is 0°.

[OR]

(iii) [Alternative option] Cartesian equation of line AB and verification that P(5, 10, 7) lies on it:

The first corridor passes through A(1, 2, −1) with direction ratios 1 : 2 : 2 (from part (i)).

Cartesian equation of line AB:
(x − 1)/1 = (y − 2)/2 = (z − (−1))/2

∴ Cartesian equation of the first flight corridor is (x − 1)/1 = (y − 2)/2 = (z + 1)/2.

Verification for P(5, 10, 7):
Substituting x = 5, y = 10, z = 7:
(5 − 1)/1 = 4/1 = 4
(10 − 2)/2 = 8/2 = 4
(7 + 1)/2 = 8/2 = 4

∵ All three ratios are equal (= 4), point P(5, 10, 7) lies on the first flight corridor.

∴ The cartesian equation of the first flight corridor is (x − 1)/1 = (y − 2)/2 = (z + 1)/2, and P(5, 10, 7) lies on it.

Question 3

A drone delivery company uses a coordinate system (in km) to track its drones. Drone A is stationed at the point P(1, −1, 2) and flies in the direction parallel to the line  (x − 3)/2 = (y + 1)/(−1) = (z − 2)/3.  Drone B is stationed at the point Q(2, 1, −1) and flies in the direction parallel to the line  (x − 1)/3 = (y − 2)/2 = (z + 1)/(−2).

Based on the above information, answer the following:
(i)  Write the direction ratios of the path of Drone A.  [1 mark]
(ii) Write the vector equation of the path of Drone B.  [1 mark]
(iii) Find the angle between the paths of Drone A and Drone B, OR
(iii) Determine whether the paths of Drone A and Drone B are parallel, intersecting, or skew lines.  [2 marks]

Accepted Answer

(i) Direction ratios of the path of Drone A:

The path of Drone A is parallel to the line (x − 3)/2 = (y + 1)/(−1) = (z − 2)/3.

∴ Direction ratios of the path of Drone A are 2, −1, 3.

(ii) Vector equation of the path of Drone B:

Drone B passes through Q(2, 1, −1) and its path is parallel to the direction vector b⃗₂ = 3î + 2ĵ − 2k̂.

∴ Vector equation of the path of Drone B is:

r⃗ = (2î + ĵ − k̂) + λ(3î + 2ĵ − 2k̂),  λ ∈ ℝ

(iii) [Main option] Finding the angle between the paths of Drone A and Drone B:

Direction vector of path of Drone A: b⃗₁ = 2î − ĵ + 3k̂

Direction vector of path of Drone B: b⃗₂ = 3î + 2ĵ − 2k̂

Using the formula cosθ = |b⃗₁ · b⃗₂| / (|b⃗₁| |b⃗₂|),

b⃗₁ · b⃗₂ = (2)(3) + (−1)(2) + (3)(−2) = 6 − 2 − 6 = −2

|b⃗₁| = √(2² + (−1)² + 3²) = √(4 + 1 + 9) = √14

|b⃗₂| = √(3² + 2² + (−2)²) = √(9 + 4 + 4) = √17

cosθ = |−2| / (√14 · √17) = 2/√238

∴ Angle between the paths of Drone A and Drone B = θ = cos⁻¹(2/√238)

[OR option] Determining whether the paths are parallel, intersecting, or skew:

Path of Drone A:  r⃗ = (î − ĵ + 2k̂) + μ(2î − ĵ + 3k̂),  μ ∈ ℝ
Path of Drone B:  r⃗ = (2î + ĵ − k̂) + λ(3î + 2ĵ − 2k̂),  λ ∈ ℝ

Step 1 — Check for parallel paths:

b⃗₁ = 2î − ĵ + 3k̂ and b⃗₂ = 3î + 2ĵ − 2k̂

For b⃗₁ ∥ b⃗₂, we need 2/3 = −1/2 = 3/(−2), which is not satisfied.

∴ The paths are NOT parallel.

Step 2 — Check for intersection using the scalar triple product:

a⃗₂ − a⃗₁ = (2î + ĵ − k̂) − (î − ĵ + 2k̂) = î + 2ĵ − 3k̂

Compute (a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂):

b⃗₁ × b⃗₂ = |î  ĵ  k̂; 2  −1  3; 3  2  −2|
= î[(−1)(−2) − (3)(2)] − ĵ[(2)(−2) − (3)(3)] + k̂[(2)(2) − (−1)(3)]
= î[2 − 6] − ĵ[−4 − 9] + k̂[4 + 3]
= −4î + 13ĵ + 7k̂

(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂) = (1)(−4) + (2)(13) + (−3)(7)
= −4 + 26 − 21 = 1 ≠ 0

∵ The scalar triple product ≠ 0, the lines do not intersect.

∴ The paths of Drone A and Drone B are SKEW lines.

Question 4

A drone delivery system operates along two fixed flight corridors in 3D airspace. Corridor 1 passes through the point A(1, 2, 3) with direction ratios 2, 3, 4, and Corridor 2 passes through the point B(2, 1, 4) with direction ratios 1, 2, −1.

(i) Write the vector equation of each corridor. [1 mark]
(ii) Find whether the two corridors are coplanar (i.e., do the flight paths lie in a common plane). [1 mark]
(iii) A ground control station is located at point P(4, 5, 7). Find the foot of the perpendicular from P to Corridor 1, and hence find the shortest distance from the control station to Corridor 1. [2 marks]

Accepted Answer

(i) Vector equations of the two corridors:

Corridor 1: r⃗ = (î + 2ĵ + 3k̂) + λ(2î + 3ĵ + 4k̂)

Corridor 2: r⃗ = (2î + ĵ + 4k̂) + μ(î + 2ĵ − k̂)

[1 mark for both correct vector equations]

(ii) To check coplanarity of the two corridors:

The two lines are coplanar if and only if (b⃗₂ − b⃗₁) · (d⃗₁ × d⃗₂) = 0,
where b⃗₁ = î + 2ĵ + 3k̂, b⃗₂ = 2î + ĵ + 4k̂, d⃗₁ = 2î + 3ĵ + 4k̂, d⃗₂ = î + 2ĵ − k̂.

b⃗₂ − b⃗₁ = (2−1)î + (1−2)ĵ + (4−3)k̂ = î − ĵ + k̂

d⃗₁ × d⃗₂ = |î  ĵ  k̂ / 2  3  4 / 1  2  −1|

= î(3·(−1) − 4·2) − ĵ(2·(−1) − 4·1) + k̂(2·2 − 3·1)

= î(−3 − 8) − ĵ(−2 − 4) + k̂(4 − 3)

= −11î + 6ĵ + k̂

(b⃗₂ − b⃗₁) · (d⃗₁ × d⃗₂) = (1)(−11) + (−1)(6) + (1)(1) = −11 − 6 + 1 = −16 ≠ 0

∴ The two corridors are NOT coplanar, i.e., the flight paths do not lie in a common plane.

[1 mark]

(iii) Finding the foot of perpendicular from P(4, 5, 7) to Corridor 1:

A general point Q on Corridor 1 is:
Q = (1 + 2λ, 2 + 3λ, 3 + 4λ)

PQ⃗ = Q − P = (1 + 2λ − 4)î + (2 + 3λ − 5)ĵ + (3 + 4λ − 7)k̂
= (2λ − 3)î + (3λ − 3)ĵ + (4λ − 4)k̂

For Q to be the foot of perpendicular, PQ⃗ · d⃗₁ = 0:

(2λ − 3)(2) + (3λ − 3)(3) + (4λ − 4)(4) = 0

4λ − 6 + 9λ − 9 + 16λ − 16 = 0

29λ − 31 = 0

⟹ λ = 31/29

∴ Foot of perpendicular Q = (1 + 62/29, 2 + 93/29, 3 + 124/29)
= (91/29, 151/29, 211/29)

Shortest distance = |PQ⃗|:

PQ⃗ = (2·(31/29) − 3)î + (3·(31/29) − 3)ĵ + (4·(31/29) − 4)k̂
= (62/29 − 87/29)î + (93/29 − 87/29)ĵ + (124/29 − 116/29)k̂
= (−25/29)î + (6/29)ĵ + (8/29)k̂

|PQ⃗| = (1/29)√(625 + 36 + 64) = (1/29)√725 = (1/29)·5√29 = 5/√29 = 5√29/29

∴ The foot of the perpendicular from the ground control station to Corridor 1 is Q(91/29, 151/29, 211/29), and the shortest distance from P to Corridor 1 is 5√29/29 units.

[1 mark for finding λ and foot of perpendicular; 1 mark for computing the shortest distance]

Question 5

A drone delivery company plans flight paths using 3D coordinates (in km). Two flight corridors are modelled as straight lines:

Corridor 1 passes through the control tower at A(1, 2, −1) with direction ratios 2, 3, 1.
Corridor 2 passes through the relay station at B(3, 0, 1) with direction ratios 1, −1, 2.

Based on this information, answer the following:

(i) Write the vector equation of Corridor 1.
(ii) Write the cartesian equation of Corridor 2.
(iii) Find the angle between the two flight corridors. OR
(iii) Find the shortest distance between the two flight corridors.

Accepted Answer

(i) The vector equation of Corridor 1 is of the form r⃗ = a⃗ + λb⃗, where a⃗ is the position vector of point A and b⃗ is the direction vector.

Here, a⃗ = î + 2ĵ − k̂ and b⃗ = 2î + 3ĵ + k̂.

∴ Vector equation of Corridor 1: r⃗ = (î + 2ĵ − k̂) + λ(2î + 3ĵ + k̂)  [1 mark]

(ii) The cartesian equation of a line through (x₁, y₁, z₁) with direction ratios a, b, c is:
(x − x₁)/a = (y − y₁)/b = (z − z₁)/c

Here, (x₁, y₁, z₁) = (3, 0, 1) and direction ratios = 1, −1, 2.

∴ Cartesian equation of Corridor 2: (x − 3)/1 = (y − 0)/(−1) = (z − 1)/2  [1 mark]

(iii) The angle θ between two lines with direction vectors b⃗₁ and b⃗₂ is given by:
cosθ = |b⃗₁ · b⃗₂| / (|b⃗₁| |b⃗₂|)

Here, b⃗₁ = 2î + 3ĵ + k̂ and b⃗₂ = î − ĵ + 2k̂.

b⃗₁ · b⃗₂ = (2)(1) + (3)(−1) + (1)(2) = 2 − 3 + 2 = 1

|b⃗₁| = √(4 + 9 + 1) = √14
|b⃗₂| = √(1 + 1 + 4) = √6

cosθ = |1| / (√14 · √6) = 1/√84 = 1/(2√21)

∴ θ = cos⁻¹(1/(2√21))  [2 marks]

OR

(iii) The shortest distance between two skew lines r⃗ = a⃗₁ + λb⃗₁ and r⃗ = a⃗₂ + μb⃗₂ is:
SD = |(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂)| / |b⃗₁ × b⃗₂|

Here, a⃗₁ = î + 2ĵ − k̂,  a⃗₂ = 3î + 0ĵ + k̂,  b⃗₁ = 2î + 3ĵ + k̂,  b⃗₂ = î − ĵ + 2k̂.

a⃗₂ − a⃗₁ = (3−1)î + (0−2)ĵ + (1−(−1))k̂ = 2î − 2ĵ + 2k̂

b⃗₁ × b⃗₂ = |î  ĵ  k̂ |
              |2  3  1 |
              |1 −1  2 |

= î[(3)(2) − (1)(−1)] − ĵ[(2)(2) − (1)(1)] + k̂[(2)(−1) − (3)(1)]
= î[6 + 1] − ĵ[4 − 1] + k̂[−2 − 3]
= 7î − 3ĵ − 5k̂

|b⃗₁ × b⃗₂| = √(49 + 9 + 25) = √83

(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂) = (2)(7) + (−2)(−3) + (2)(−5)
= 14 + 6 − 10 = 10

∴ SD = |10| / √83 = 10/√83 km  [2 marks]

Question 6

A drone delivery system operates along fixed flight corridors modelled as straight lines in 3D space. Two drones are assigned the following flight paths:

Drone A travels along the line: r⃗ = (î + 2ĵ − k̂) + λ(2î − ĵ + 3k̂)

Drone B travels along the line: r⃗ = (3î + ĵ + 5k̂) + μ(î + 2ĵ − k̂)

A flight controller needs to determine whether the two corridors intersect (collision risk), are parallel, or are skew lines (safe). Based on your analysis:

(i) Show that the direction vectors of the two flight paths are NOT parallel. [1]

(ii) Check whether the lines intersect or are skew by testing the consistency of the system formed when the general points on both lines are equated. [1]

(iii) Find the shortest distance between the two flight corridors, and hence determine whether the drones are at risk of collision. [2]

Accepted Answer

(i) [1 mark]

The direction vector of Drone A's path is b⃗₁ = 2î − ĵ + 3k̂.
The direction vector of Drone B's path is b⃗₂ = î + 2ĵ − k̂.

For the lines to be parallel, we need b⃗₁ = k·b⃗₂ for some scalar k, i.e.,

2/1 = −1/2 = 3/(−1)

⟹ 2 ≠ −1/2

∴ The direction vectors are NOT parallel, so the two flight corridors are not parallel.

(ii) [1 mark]

The position vectors of the given points on the lines are:
a⃗₁ = î + 2ĵ − k̂,   a⃗₂ = 3î + ĵ + 5k̂.

Equating the general points on the two lines:
î + 2ĵ − k̂ + λ(2î − ĵ + 3k̂) = 3î + ĵ + 5k̂ + μ(î + 2ĵ − k̂)

On comparing coefficients of î, ĵ, k̂:

1 + 2λ = 3 + μ   ⟹   2λ − μ = 2   ...(i)
2 − λ = 1 + 2μ   ⟹   −λ − 2μ = −1   ⟹   λ + 2μ = 1   ...(ii)
−1 + 3λ = 5 − μ   ⟹   3λ + μ = 6   ...(iii)

From (i): μ = 2λ − 2. Substituting in (ii):
λ + 2(2λ − 2) = 1   ⟹   λ + 4λ − 4 = 1   ⟹   5λ = 5   ⟹   λ = 1, μ = 0.

Checking in (iii): 3(1) + (0) = 3 ≠ 6.

∵ The values λ = 1, μ = 0 do NOT satisfy equation (iii), the system is inconsistent.

∴ The two flight corridors are SKEW lines (neither parallel nor intersecting).

(iii) [2 marks]

The shortest distance between two skew lines is given by:

SD = |(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂)| / |b⃗₁ × b⃗₂|

Step 1: Compute a⃗₂ − a⃗₁.
a⃗₂ − a⃗₁ = (3î + ĵ + 5k̂) − (î + 2ĵ − k̂) = 2î − ĵ + 6k̂

Step 2: Compute b⃗₁ × b⃗₂.
b⃗₁ × b⃗₂ = |î  ĵ  k̂|
              |2  −1  3|
              |1   2  −1|

= î[(−1)(−1) − (3)(2)] − ĵ[(2)(−1) − (3)(1)] + k̂[(2)(2) − (−1)(1)]
= î[1 − 6] − ĵ[−2 − 3] + k̂[4 + 1]
= −5î + 5ĵ + 5k̂

Step 3: Compute |b⃗₁ × b⃗₂|.
|b⃗₁ × b⃗₂| = √((−5)² + 5² + 5²) = √(25 + 25 + 25) = √75 = 5√3

Step 4: Compute (a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂).
(2î − ĵ + 6k̂) · (−5î + 5ĵ + 5k̂)
= (2)(−5) + (−1)(5) + (6)(5)
= −10 − 5 + 30
= 15

Step 5: Apply the formula.
SD = |15| / 5√3 = 15 / 5√3 = 3/√3 = 3√3/3 = √3

∴ Shortest distance between the two flight corridors = √3 units.

∵ The shortest distance √3 ≠ 0, the two corridors do not intersect.
∴ There is NO collision risk — the drones are safe, as the flight paths are skew lines separated by a shortest distance of √3 units.

Question 7

A drone surveillance system monitors a restricted zone. The flight path of Drone A is modelled by the line L₁ whose vector equation is r⃗ = (î + 2ĵ − k̂) + λ(2î − ĵ + 3k̂), and the flight path of Drone B is modelled by the line L₂ whose vector equation is r⃗ = (3î + ĵ + 2k̂) + μ(î + 2ĵ − k̂).

Based on this information, answer the following:

(i) Write the Cartesian equations of the flight path L₁ of Drone A.  [1 mark]

(ii) Find the direction cosines of the flight path L₂ of Drone B.  [1 mark]

(iii) Determine whether the two flight paths L₁ and L₂ are parallel, intersecting, or skew. If they are skew lines, find the shortest distance between them.  [2 marks]

Accepted Answer

(i) The vector equation of L₁ is r⃗ = (î + 2ĵ − k̂) + λ(2î − ĵ + 3k̂).

Here, the line passes through the point (1, 2, −1) and has direction ratios 2, −1, 3.

The Cartesian equation of L₁ is:

(x − 1)/2 = (y − 2)/(−1) = (z + 1)/3

(ii) The direction vector of L₂ is b⃗₂ = î + 2ĵ − k̂.

|b⃗₂| = √(1² + 2² + (−1)²) = √(1 + 4 + 1) = √6

∴ Direction cosines of L₂ are: 1/√6, 2/√6, −1/√6.

(iii) The direction vector of L₁ is b⃗₁ = 2î − ĵ + 3k̂.

The direction vector of L₂ is b⃗₂ = î + 2ĵ − k̂.

Check for parallelism: b⃗₁ = λb⃗₂ would require 2/1 = −1/2 = 3/(−1), i.e., 2 = −1/2 = −3, which is false.
∴ L₁ and L₂ are not parallel.

Check for intersection: Suppose the lines meet. Then for some λ and μ,

(1 + 2λ) = (3 + μ)   ...(i)
(2 − λ) = (1 + 2μ)   ...(ii)
(−1 + 3λ) = (2 − μ)  ...(iii)

From (i): μ = 2λ − 2   ...(iv)

Substituting (iv) in (ii): 2 − λ = 1 + 2(2λ − 2) = 1 + 4λ − 4 = 4λ − 3
⟹ 5 = 5λ ⟹ λ = 1 and μ = 2(1) − 2 = 0.

Verifying in (iii): L.H.S. = −1 + 3(1) = 2; R.H.S. = 2 − 0 = 2. ✓

Since equation (iii) is satisfied, the lines are consistent.

∵ All three equations are satisfied, L₁ and L₂ are intersecting lines.

∴ The two flight paths L₁ and L₂ intersect each other (they are neither parallel nor skew). The shortest distance between them is 0.

Question 8

A drone delivery company plans flight paths for two drones operating in a warehouse modelled using a 3D coordinate system (units in metres).

Drone 1 starts at point A(1, −1, 2) and flies in the direction of the line (x − 3)/2 = (y + 1)/(−1) = (z − 2)/3.

Drone 2 flies along the path r⃗ = (2î + ĵ − k̂) + μ(î + 2ĵ − k̂).

The warehouse manager needs to ensure the two flight paths do NOT intersect and must calculate the minimum clearance (shortest distance) between the paths to check for collision risk.

(i) Write the vector equation of the flight path of Drone 1.  [1]
(ii) Write the direction vectors b⃗₁ and b⃗₂ of the two paths and verify that the paths are skew lines (not parallel).  [1]
(iii) Calculate the shortest distance between the two flight paths and state whether a collision risk exists if the safety clearance required is 2 metres.  [2]
       OR
(iii) Find the vector b⃗₁ × b⃗₂ and hence find the unit vector perpendicular to both flight paths.  [2]

Accepted Answer

(i) The direction of Drone 1's path is given by the cartesian line (x − 3)/2 = (y + 1)/(−1) = (z − 2)/3, so its direction vector is b⃗₁ = 2î − ĵ + 3k̂.

Drone 1 passes through A(1, −1, 2), so its position vector is a⃗₁ = î − ĵ + 2k̂.

∴ Vector equation of flight path of Drone 1 is:
r⃗ = (î − ĵ + 2k̂) + λ(2î − ĵ + 3k̂)

(ii) From part (i): b⃗₁ = 2î − ĵ + 3k̂

From Drone 2's equation r⃗ = (2î + ĵ − k̂) + μ(î + 2ĵ − k̂): b⃗₂ = î + 2ĵ − k̂

For the paths to be parallel, b⃗₁ = kb⃗₂ for some scalar k.
Comparing components: 2/1 = −1/2 — this gives 2 ≠ −1/2.

∴ b⃗₁ is not a scalar multiple of b⃗₂, so the two flight paths are NOT parallel.

(iii) The shortest distance formula for skew lines is:

S.D. = |(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂)| / |b⃗₁ × b⃗₂|

Here, a⃗₁ = î − ĵ + 2k̂  and  a⃗₂ = 2î + ĵ − k̂

⟹ a⃗₂ − a⃗₁ = (2−1)î + (1−(−1))ĵ + (−1−2)k̂ = î + 2ĵ − 3k̂

Now computing b⃗₁ × b⃗₂:

b⃗₁ × b⃗₂ = |î  ĵ  k̂ |
              |2  −1  3 |
              |1   2  −1|

= î[(−1)(−1) − (3)(2)] − ĵ[(2)(−1) − (3)(1)] + k̂[(2)(2) − (−1)(1)]

= î[1 − 6] − ĵ[−2 − 3] + k̂[4 + 1]

= −5î + 5ĵ + 5k̂

|b⃗₁ × b⃗₂| = √((−5)² + 5² + 5²) = √(25 + 25 + 25) = √75 = 5√3

(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂) = (î + 2ĵ − 3k̂) · (−5î + 5ĵ + 5k̂)
= (1)(−5) + (2)(5) + (−3)(5)
= −5 + 10 − 15
= −10

∴ S.D. = |−10| / 5√3 = 10 / 5√3 = 2/√3 = 2√3/3 metres

Since 2√3/3 ≈ 1.15 metres < 2 metres (required safety clearance),

∴ Shortest distance = 2√3/3 metres ≈ 1.15 m. Since this is less than the required 2 m clearance, a collision risk EXISTS.

OR

(iii) From above working:
b⃗₁ × b⃗₂ = −5î + 5ĵ + 5k̂

|b⃗₁ × b⃗₂| = √((−5)² + 5² + 5²) = √(25 + 25 + 25) = √75 = 5√3

The unit vector perpendicular to both flight paths is:

n̂ = (b⃗₁ × b⃗₂) / |b⃗₁ × b⃗₂|

= (−5î + 5ĵ + 5k̂) / 5√3

= (−î + ĵ + k̂) / √3

∴ Required unit vector perpendicular to both paths = (1/√3)(−î + ĵ + k̂)

Question 9

Assertion (A): The lines whose direction ratios are 2, −1, 3 and 4, −2, 6 are parallel to each other.

Reason (R): Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are parallel if and only if a₁/a₂ = b₁/b₂ = c₁/c₂.

Accepted Answer

Option (a) is correct.

Explanation: The condition for two lines to be parallel is a₁/a₂ = b₁/b₂ = c₁/c₂, which is exactly what Reason (R) states — so R is true.

Checking Assertion (A): direction ratios of the two lines are 2, −1, 3 and 4, −2, 6.
2/4 = −1/(−2) = 3/6 = 1/2.
∵ the ratios are equal, the lines are parallel — so A is true, and R is the correct explanation of A.

Question 10

The direction cosines of a line which makes equal angles with the positive directions of the coordinate axes are:

Accepted Answer

Option (A) is correct.

Explanation: For a line making equal angles α with all three positive coordinate axes, l = m = n = cosα.

∵ cos²α + cos²α + cos²α = 1 ⟹ 3cos²α = 1 ⟹ cosα = 1/√3.

∴ The direction cosines are 1/√3, 1/√3, 1/√3.

Question 11

Find the direction cosines of the line whose direction ratios are 2, −3, 6.

Accepted Answer

Option (A) is correct.

Explanation: If direction ratios are a, b, c then direction cosines are l = a/√(a² + b² + c²), m = b/√(a² + b² + c²), n = c/√(a² + b² + c²).

Here a = 2, b = −3, c = 6, so √(a² + b² + c²) = √(4 + 9 + 36) = √49 = 7.

∴ Direction cosines are 2/7, −3/7, 6/7.

Question 12

Find the coordinates of the foot of the perpendicular drawn from the point P(1, 2, 3) to the line  (x − 1)/2 = (y + 1)/1 = (z − 2)/3.

Accepted Answer

Let the foot of the perpendicular from P(1, 2, 3) to the line be Q.

A general point on the line  (x − 1)/2 = (y + 1)/1 = (z − 2)/3 = λ  is

Q = (1 + 2λ,  −1 + λ,  2 + 3λ).

Then  PQ⃗ = (1 + 2λ − 1)î + (−1 + λ − 2)ĵ + (2 + 3λ − 3)k̂
         = 2λ î + (λ − 3)ĵ + (3λ − 1)k̂.

The direction vector of the line is b⃗ = 2î + ĵ + 3k̂.

For the foot of the perpendicular,  PQ⃗ · b⃗ = 0:

(2λ)(2) + (λ − 3)(1) + (3λ − 1)(3) = 0

⟹ 4λ + λ − 3 + 9λ − 3 = 0

⟹ 14λ − 6 = 0  ⟹  λ = 3/7.

∴ Foot of perpendicular = (1 + 2·(3/7),  −1 + 3/7,  2 + 3·(3/7))
                        = (1 + 6/7,  −1 + 3/7,  2 + 9/7)
                        = (13/7,  −4/7,  23/7).

Question 13

Find the direction cosines of the line whose direction ratios are 2, −3, 6. Also write the vector equation of the line passing through the point (1, −2, 4) and having the above direction ratios.

Accepted Answer

The direction ratios are a = 2, b = −3, c = 6.

√(a² + b² + c²) = √(4 + 9 + 36) = √49 = 7

∴ Direction cosines are l = 2/7, m = −3/7, n = 6/7.

The line passes through the point (1, −2, 4), so position vector a⃗ = î − 2ĵ + 4k̂, and direction vector b⃗ = 2î − 3ĵ + 6k̂.

∴ Required vector equation of the line is r⃗ = (î − 2ĵ + 4k̂) + λ(2î − 3ĵ + 6k̂), where λ ∈ ℝ.

Question 14

Find the distance of the point (2, −1, 3) from the plane 3x − 4y + 12z = 5.

Accepted Answer

The distance of a point (x₁, y₁, z₁) from the plane ax + by + cz = d is given by

D = |ax₁ + by₁ + cz₁ − d| / √(a² + b² + c²)

Here (x₁, y₁, z₁) = (2, −1, 3) and the plane is 3x − 4y + 12z − 5 = 0, so a = 3, b = −4, c = 12, d = 5.

D = |3(2) + (−4)(−1) + 12(3) − 5| / √(3² + (−4)² + 12²)
      = |6 + 4 + 36 − 5| / √(9 + 16 + 144)
      = |41| / √169
      = 41 / 13

∴ Required distance = 41/13 units.

Question 15

Find the angle between the lines whose direction ratios are proportional to (1, −2, 2) and (3, 6, 2). Also find the direction cosines of a line perpendicular to both of them.

Accepted Answer

Let b⃗₁ = î − 2ĵ + 2k̂ and b⃗₂ = 3î + 6ĵ + 2k̂ be the direction vectors of the two lines.

|b⃗₁| = √(1² + (−2)² + 2²) = √(1 + 4 + 4) = √9 = 3

|b⃗₂| = √(3² + 6² + 2²) = √(9 + 36 + 4) = √49 = 7

b⃗₁ · b⃗₂ = (1)(3) + (−2)(6) + (2)(2) = 3 − 12 + 4 = −5

The angle θ between the two lines is given by:

cosθ = |b⃗₁ · b⃗₂| / (|b⃗₁| |b⃗₂|) = |−5| / (3 × 7) = 5/21

∴ θ = cos⁻¹(5/21)

Now, a line perpendicular to both b⃗₁ and b⃗₂ has direction along b⃗₁ × b⃗₂.

b⃗₁ × b⃗₂ = |  î   ĵ   k̂  |
             |  1  −2   2  |
             |  3   6   2  |

= î[(−2)(2) − (2)(6)] − ĵ[(1)(2) − (2)(3)] + k̂[(1)(6) − (−2)(3)]

= î[−4 − 12] − ĵ[2 − 6] + k̂[6 + 6]

= −16î + 4ĵ + 12k̂

|b⃗₁ × b⃗₂| = √((−16)² + 4² + 12²) = √(256 + 16 + 144) = √416 = 4√26

The direction cosines of the line perpendicular to both are:

l = −16/(4√26) = −4/√26,   m = 4/(4√26) = 1/√26,   n = 12/(4√26) = 3/√26

∴ The angle between the lines is cos⁻¹(5/21) and the direction cosines of the line perpendicular to both are (−4/√26, 1/√26, 3/√26).

Question 16

Find the vector and cartesian equations of the line passing through the point (1, −2, 3) and perpendicular to the lines whose direction ratios are (2, 3, −1) and (1, −2, 3).

Accepted Answer

The direction vector b⃗ of the required line is perpendicular to both the given lines, so b⃗ = d⃗₁ × d⃗₂, where d⃗₁ = 2î + 3ĵ − k̂ and d⃗₂ = î − 2ĵ + 3k̂.

b⃗ = d⃗₁ × d⃗₂ = |î  ĵ  k̂ |
                   |2  3  −1|
                   |1 −2   3|

b⃗ = î[(3)(3) − (−1)(−2)] − ĵ[(2)(3) − (−1)(1)] + k̂[(2)(−2) − (3)(1)]

b⃗ = î[9 − 2] − ĵ[6 + 1] + k̂[−4 − 3]

b⃗ = 7î − 7ĵ − 7k̂

We may take b⃗ = î − ĵ − k̂  (dividing by 7).

The line passes through the point A(1, −2, 3), so a⃗ = î − 2ĵ + 3k̂.

Vector equation:
r⃗ = (î − 2ĵ + 3k̂) + λ(î − ĵ − k̂),  λ ∈ ℝ

Cartesian equation:
(x − 1)/1 = (y + 2)/(−1) = (z − 3)/(−1)

Question 17

Find the shortest distance between the following lines:

$\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$

$\vec{r} = (4\hat{i} + 3\hat{j} + \hat{k}) + \mu(2\hat{i} + 4\hat{j} - 2\hat{k})$

Accepted Answer

Here, a⃗₁ = 2î − ĵ + 3k̂, b⃗₁ = î + 2ĵ − k̂

and a⃗₂ = 4î + 3ĵ + k̂, b⃗₂ = 2î + 4ĵ − 2k̂

Observe that b⃗₂ = 2b⃗₁, so the two lines are parallel.

For parallel lines, the shortest distance is given by:

SD = |( a⃗₂ − a⃗₁ ) × b̂₁| = |(a⃗₂ − a⃗₁) × b⃗₁| / |b⃗₁|

Now, a⃗₂ − a⃗₁ = (4 − 2)î + (3 − (−1))ĵ + (1 − 3)k̂ = 2î + 4ĵ − 2k̂

(a⃗₂ − a⃗₁) × b⃗₁ = |  î    ĵ    k̂  |
                       |  2    4   −2  |
                       |  1    2   −1  |

= î[(4)(−1) − (−2)(2)] − ĵ[(2)(−1) − (−2)(1)] + k̂[(2)(2) − (4)(1)]

= î[−4 + 4] − ĵ[−2 + 2] + k̂[4 − 4]

= î(0) − ĵ(0) + k̂(0) = 0⃗

∵ (a⃗₂ − a⃗₁) × b⃗₁ = 0⃗, the vector (a⃗₂ − a⃗₁) is parallel to b⃗₁.

∴ The two lines are not merely parallel — they are coincident (lie on the same line).

∴ Shortest distance between the two lines = 0 units.

Question 18

Find the vector and cartesian equations of the line passing through the point (1, 2, −1) and perpendicular to both the lines whose direction ratios are (2, 1, −3) and (1, −2, 1).

Accepted Answer

The direction vector b⃗ of the required line is perpendicular to both the given direction vectors.

Let d⃗₁ = 2î + ĵ − 3k̂ and d⃗₂ = î − 2ĵ + k̂.

Then b⃗ = d⃗₁ × d⃗₂.

Computing the cross product:

b⃗ = |  î    ĵ    k̂  |
     |  2    1   −3  |
     |  1   −2    1  |

b⃗ = î[(1)(1) − (−3)(−2)] − ĵ[(2)(1) − (−3)(1)] + k̂[(2)(−2) − (1)(1)]

b⃗ = î[1 − 6] − ĵ[2 + 3] + k̂[−4 − 1]

b⃗ = −5î − 5ĵ − 5k̂

We may take b⃗ = î + ĵ + k̂   (dividing by −5, direction ratios (1, 1, 1)).

The line passes through the point A(1, 2, −1), so a⃗ = î + 2ĵ − k̂.

Vector equation of the required line:

r⃗ = (î + 2ĵ − k̂) + λ(î + ĵ + k̂),   λ ∈ ℝ

Cartesian equation of the required line:

(x − 1)/1 = (y − 2)/1 = (z + 1)/1

Question 19

Find the foot of the perpendicular drawn from the point P(2, 3, −1) to the line

(x − 1)/2 = (y + 2)/3 = (z − 3)/(−1).

Also find the length of the perpendicular.

Accepted Answer

Let the foot of the perpendicular from P(2, 3, −1) on the given line be Q.

The parametric form of the line (x − 1)/2 = (y + 2)/3 = (z − 3)/(−1) = λ gives the general point

Q = (1 + 2λ, −2 + 3λ, 3 − λ).

The direction ratios of PQ are:

PQ⃗ = Q − P = (1 + 2λ − 2, −2 + 3λ − 3, 3 − λ − (−1))
     = (2λ − 1, 3λ − 5, 4 − λ).

The direction ratios of the given line are (2, 3, −1).

For PQ⃗ ⊥ line, we need PQ⃗ · b⃗ = 0:

2(2λ − 1) + 3(3λ − 5) + (−1)(4 − λ) = 0

⟹ 4λ − 2 + 9λ − 15 − 4 + λ = 0

⟹ 14λ − 21 = 0

⟹ λ = 3/2.

Substituting λ = 3/2:

Q = (1 + 2·(3/2), −2 + 3·(3/2), 3 − 3/2)
  = (1 + 3, −2 + 9/2, 3/2)
  = (4, 5/2, 3/2).

∴ Foot of the perpendicular = (4, 5/2, 3/2).

Length of the perpendicular = |PQ⃗|:

PQ⃗ = (4 − 2, 5/2 − 3, 3/2 − (−1))
    = (2, −1/2, 5/2).

|PQ⃗| = √(2² + (−1/2)² + (5/2)²)
      = √(4 + 1/4 + 25/4)
      = √(16/4 + 1/4 + 25/4)
      = √(42/4)
      = √42 / 2.

∴ Length of the perpendicular = √42/2 units.

Question 20

Find the angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l² + m² − n² = 0.

Accepted Answer

From the first equation: l + m + n = 0  ⟹  n = −(l + m)  ...(i)

Substituting into the second equation l² + m² − n² = 0:

l² + m² − (l + m)² = 0

⟹  l² + m² − l² − 2lm − m² = 0

⟹  −2lm = 0

⟹  lm = 0

∴ Either l = 0  or  m = 0.

Case 1: l = 0.
From (i): n = −m.  So the direction ratios are (0, m, −m), i.e., (0, 1, −1).

Case 2: m = 0.
From (i): n = −l.  So the direction ratios are (l, 0, −l), i.e., (1, 0, −1).

Let b⃗₁ = 0î + 1ĵ − 1k̂  and  b⃗₂ = 1î + 0ĵ − 1k̂.

Using the formula for the acute angle θ between two lines:

cosθ = |b⃗₁ · b⃗₂| / (|b⃗₁| |b⃗₂|)

b⃗₁ · b⃗₂ = (0)(1) + (1)(0) + (−1)(−1) = 1

|b⃗₁| = √(0² + 1² + (−1)²) = √2

|b⃗₂| = √(1² + 0² + (−1)²) = √2

∴ cosθ = |1| / (√2 · √2) = 1/2

∴ θ = cos⁻¹(1/2) = π/3

∴ The angle between the two lines is π/3 (i.e., 60°).

Three Dimensional Geometry: Class 12 Maths Practice Questions

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