Question 1

The scalar triple product of vectors **a** = **i** + 2**j** + 3**k**, **b** = 2**i** − **j** + **k**, and **c** = **i** + **j** − **k** is:

Accepted Answer

The scalar triple product **a** · (**b** × **c**) can be computed using the determinant formula: **a** · (**b** × **c**) = |a₁ a₂ a₃| |b₁ b₂ b₃| |c₁ c₂ c₃| = |1 2 3| |2 −1 1| |1 1 −1|. Expanding along the first row: = 1·|−1 1; 1 −1| − 2·|2 1; 1 −1| + 3·|2 −1; 1 1| = 1·(1 − 1) − 2·(−2 − 1) + 3·(2 + 1) = 1(0) − 2(−3) + 3(3) = 0 + 6 + 9 = 15.

Question 2

The direction cosines of a line passing through points P(1, 2, 3) and Q(4, 5, 6) are:

Accepted Answer

Direction cosines are found using the direction ratios of the line. The direction ratios are obtained by subtracting coordinates: (4−1, 5−2, 6−3) = (3, 3, 3). The magnitude is √(3² + 3² + 3²) = √27 = 3√3. Direction cosines are (3/(3√3), 3/(3√3), 3/(3√3)) = (1/√3, 1/√3, 1/√3). Rationalizing: (√3/3, √3/3, √3/3).

Question 3

The equation of the plane passing through the point (2, 3, 1) and perpendicular to the vector **n** = (1, −2, 3) is:

Accepted Answer

The equation of a plane perpendicular to vector **n** = (a, b, c) and passing through point (x₀, y₀, z₀) is: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0. Substituting **n** = (1, −2, 3) and point (2, 3, 1): 1(x − 2) − 2(y − 3) + 3(z − 1) = 0. Expanding: x − 2 − 2y + 6 + 3z − 3 = 0. Simplifying: x − 2y + 3z + 1 = 0.

Question 4

The distance from the point (1, 2, 3) to the plane 2x + 2y + z − 9 = 0 is:

Accepted Answer

The distance from point P(x₀, y₀, z₀) to plane ax + by + cz + d = 0 is given by: distance = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²). Here, P = (1, 2, 3), plane: 2x + 2y + z − 9 = 0, so a = 2, b = 2, c = 1, d = −9. Numerator: |2(1) + 2(2) + 1(3) − 9| = |2 + 4 + 3 − 9| = |0| = 0. Denominator: √(4 + 4 + 1) = √9 = 3. Distance = 0/3 = 0.

Question 5

The angle between the planes x + 2y + 2z = 5 and 2x − y + 2z = 3 is:

Accepted Answer

The angle θ between two planes with normal vectors **n₁** = (a₁, b₁, c₁) and **n₂** = (a₂, b₂, c₂) is given by: cos θ = |**n₁** · **n₂**| / (|**n₁**| |**n₂**|). For plane 1: **n₁** = (1, 2, 2). For plane 2: **n₂** = (2, −1, 2). Dot product: **n₁** · **n₂** = 1(2) + 2(−1) + 2(2) = 2 − 2 + 4 = 4. Magnitude of **n₁**: |**n₁**| = √(1 + 4 + 4) = √9 = 3. Magnitude of **n₂**: |**n₂**| = √(4 + 1 + 4) = √9 = 3. cos θ = |4| / (3 × 3) = 4/9. Therefore, θ = arccos(4/9).

Question 6

The equation of the line passing through the point (1, 2, 3) and parallel to the vector **d** = (2, −1, 3) in parametric form is:

Accepted Answer

The parametric equation of a line passing through point P(x₀, y₀, z₀) and parallel to direction vector **d** = (a, b, c) is: x = x₀ + at, y = y₀ + bt, z = z₀ + ct, where t is a parameter. Substituting P = (1, 2, 3) and **d** = (2, −1, 3): x = 1 + 2t, y = 2 − t, z = 3 + 3t.

Question 7

The direction cosines of a line passing through points A(1, 2, 3) and B(4, 5, 6) are:

Accepted Answer

Direction cosines are found using the direction ratios of the line. The direction ratios are proportional to the differences in coordinates: (4−1, 5−2, 6−3) = (3, 3, 3). To find direction cosines, we normalize by dividing by the magnitude. Magnitude = √(3² + 3² + 3²) = √27 = 3√3. Direction cosines = (3/(3√3), 3/(3√3), 3/(3√3)) = (1/√3, 1/√3, 1/√3). Rationalizing: (√3/3, √3/3, √3/3).

Question 8

The equation of the plane passing through the point (2, 3, 4) and perpendicular to the vector (1, 2, 3) is:

Accepted Answer

A plane perpendicular to vector **n** = (a, b, c) and passing through point (x₀, y₀, z₀) has equation: a(x − x₀) + b(y − y₀) + c(z − z₀) = 0. Here, **n** = (1, 2, 3) and point is (2, 3, 4). Substituting: 1(x − 2) + 2(y − 3) + 3(z − 4) = 0 ⟹ x − 2 + 2y − 6 + 3z − 12 = 0 ⟹ x + 2y + 3z − 20 = 0.

Question 9

The distance from the point P(1, 2, 3) to the plane 2x − y + 2z − 9 = 0 is:

Accepted Answer

The distance from point (x₀, y₀, z₀) to plane ax + by + cz + d = 0 is: distance = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²). Here, point P(1, 2, 3) and plane 2x − y + 2z − 9 = 0. Numerator: |2(1) − 1(2) + 2(3) − 9| = |2 − 2 + 6 − 9| = |−3| = 3. Denominator: √(2² + (−1)² + 2²) = √(4 + 1 + 4) = √9 = 3. Distance = 3/3 = 1.

Question 10

The angle between the planes x + 2y + 2z = 5 and 2x − y + 2z = 8 is:

Accepted Answer

The angle θ between two planes with normal vectors **n₁** = (a₁, b₁, c₁) and **n₂** = (a₂, b₂, c₂) is given by: cos θ = |**n₁** · **n₂**| / (||**n₁**|| ||**n₂**||). For plane 1: **n₁** = (1, 2, 2). For plane 2: **n₂** = (2, −1, 2). Dot product: **n₁** · **n₂** = 1(2) + 2(−1) + 2(2) = 2 − 2 + 4 = 4. Magnitude of **n₁**: ||**n₁**|| = √(1 + 4 + 4) = √9 = 3. Magnitude of **n₂**: ||**n₂**|| = √(4 + 1 + 4) = √9 = 3. cos θ = |4| / (3 × 3) = 4/9. Therefore, θ = arccos(4/9).

Question 11

The vector equation of the line passing through points A(1, 0, 2) and B(3, 4, 6) is:

Accepted Answer

The vector equation of a line passing through point A with direction vector **d** is: **r** = **a** + t**d**, where **a** is the position vector of A and **d** is the direction vector. The direction vector is **AB** = B − A = (3−1, 4−0, 6−2) = (2, 4, 4). The position vector of A is **a** = (1, 0, 2). Therefore, the vector equation is: **r** = (1, 0, 2) + t(2, 4, 4), which can also be written as **r** = (1, 0, 2) + t(1, 2, 2) after factoring out 2 from the direction vector (both forms are equivalent).

Question 12

The direction cosines of a line are proportional to 2, −3, 6. What are the actual direction cosines of this line?

Accepted Answer

Direction cosines l, m, n of a line satisfy l² + m² + n² = 1. Given that direction ratios are proportional to 2, −3, 6, we write l : m : n = 2 : −3 : 6. Let l = 2k, m = −3k, n = 6k for some constant k. Substituting into the normalization condition: (2k)² + (−3k)² + (6k)² = 1 → 4k² + 9k² + 36k² = 1 → 49k² = 1 → k = ±1/7. Taking the positive value k = 1/7, we get l = 2/7, m = −3/7, n = 6/7. Verification: (2/7)² + (−3/7)² + (6/7)² = 4/49 + 9/49 + 36/49 = 49/49 = 1 ✓

Question 13

Find the angle between the two lines with direction ratios 1, 2, 2 and 2, 3, −6.

Accepted Answer

The angle θ between two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) is given by cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁² + b₁² + c₁²) · √(a₂² + b₂² + c₂²)). Here, (a₁, b₁, c₁) = (1, 2, 2) and (a₂, b₂, c₂) = (2, 3, −6). Numerator (dot product): 1(2) + 2(3) + 2(−6) = 2 + 6 − 12 = −4, so |−4| = 4. Denominator: √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3, and √(2² + 3² + (−6)²) = √(4 + 9 + 36) = √49 = 7. Thus cos θ = 4 / (3 · 7) = 4/21. Therefore θ = arccos(4/21) ≈ 78.96° or approximately 79°.

Question 14

The equation of a plane passing through the point (1, 2, 3) and perpendicular to the vector 2**i** + 3**j** − **k** is:

Accepted Answer

The equation of a plane perpendicular to a vector **n** = (a, b, c) and passing through point (x₀, y₀, z₀) is a(x − x₀) + b(y − y₀) + c(z − z₀) = 0. Here, **n** = (2, 3, −1) and the point is (1, 2, 3). Substituting: 2(x − 1) + 3(y − 2) − 1(z − 3) = 0 → 2x − 2 + 3y − 6 − z + 3 = 0 → 2x + 3y − z − 5 = 0. Verification: Substitute (1, 2, 3): 2(1) + 3(2) − 3 − 5 = 2 + 6 − 3 − 5 = 0 ✓

Question 15

The distance from the point (2, 3, 4) to the plane x + 2y + 2z − 9 = 0 is:

Accepted Answer

The distance from a point (x₀, y₀, z₀) to the plane ax + by + cz + d = 0 is given by the formula: distance = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²). Here, the point is (2, 3, 4) and the plane is x + 2y + 2z − 9 = 0, so a = 1, b = 2, c = 2, d = −9. Numerator: |1(2) + 2(3) + 2(4) − 9| = |2 + 6 + 8 − 9| = |7| = 7. Denominator: √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3. Distance = 7/3.

Question 16

The distance from the point P(2, 3, 4) to the line passing through A(1, 2, 3) with direction vector **d** = (1, 0, 1) is:

Accepted Answer

The distance from point P to a line through point A with direction **d** is:
distance = |(**AP**) × **d**| / |**d**|

Compute **AP** = P − A = (2, 3, 4) − (1, 2, 3) = (1, 1, 1)

Compute **AP** × **d**:
**AP** × **d** = (1, 1, 1) × (1, 0, 1)
= |**i**  **j**  **k**|
  |1    1    1|
  |1    0    1|
= **i**(1·1 − 1·0) − **j**(1·1 − 1·1) + **k**(1·0 − 1·1)
= **i**(1) − **j**(0) + **k**(−1)
= (1, 0, −1)

|**AP** × **d**| = √(1² + 0² + (−1)²) = √2
|**d**| = √(1² + 0² + 1²) = √2

distance = √2 / √2 = 1

Question 17

The equation of the plane containing the line **r** = (1, 0, 2) + t(1, 1, 1) and the point Q(2, 1, 3) is:

Accepted Answer

A plane containing a line and an external point requires two vectors in the plane. Using point P(1, 0, 2) on the line and direction **d** = (1, 1, 1), plus external point Q(2, 1, 4):

**PQ** = (1, 1, 2)

Normal vector **n** = **d** × **PQ** = (1, 1, 1) × (1, 1, 2) = (1, −1, 0)

Plane equation through P with normal **n**:
1(x − 1) − 1(y − 0) + 0(z − 2) = 0
x − y = 1

Question 18

The equation of the plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratios 2 : 3 : 4 is:

Accepted Answer

A plane perpendicular to a line has the direction ratios of the line as its normal vector. If the normal vector is (2, 3, 4) and the plane passes through point (1, 2, 3), the equation of the plane is:

2(x − 1) + 3(y − 2) + 4(z − 3) = 0

Expanding:
2x − 2 + 3y − 6 + 4z − 12 = 0
2x + 3y + 4z − 20 = 0

Verification: Substituting (1, 2, 3): 2(1) + 3(2) + 4(3) = 2 + 6 + 12 = 20 ✓

Question 19

The angle between the planes x + 2y + 2z = 5 and 2x − 3y + 6z = 7 is θ. Then cos θ equals:

Accepted Answer

The angle between two planes with normal vectors n₁ = (a₁, b₁, c₁) and n₂ = (a₂, b₂, c₂) is given by:

cos θ = |n₁ · n₂| / (|n₁| |n₂|)

For plane 1: n₁ = (1, 2, 2)
For plane 2: n₂ = (2, −3, 6)

Dot product: n₁ · n₂ = 1(2) + 2(−3) + 2(6) = 2 − 6 + 12 = 8

|n₁| = √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3
|n₂| = √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49 = 7

cos θ = |8| / (3 × 7) = 8/21

Question 20

The distance from the point (2, 3, 4) to the plane 2x − y + 2z − 9 = 0 is:

Accepted Answer

The distance from a point (x₀, y₀, z₀) to the plane ax + by + cz + d = 0 is given by:

Distance = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²)

Given: point (2, 3, 4) and plane 2x − y + 2z − 9 = 0
Here: a = 2, b = −1, c = 2, d = −9

Substitute:
Distance = |2(2) + (−1)(3) + 2(4) + (−9)| / √(2² + (−1)² + 2²)
         = |4 − 3 + 8 − 9| / √(4 + 1 + 4)
         = |0| / √9
         = 0 / 3
         = 0

The point lies on the plane.

AFCAT 3D Geometry

3D Geometry— Rules & Concept

Key Points to Remember

Exam-Specific Tips

3D Geometry — Practice Questions

60-Second Revision — 3D Geometry