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Vectors and Three D Geometry Basic Questions for AGNIVEER NAVY

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Why this topic matters · 8 min read
Vectors and 3D geometry appear in Agniveer Navy SSR/MR maths as 8-12% weightage, typically 2-3 questions. Exams test position vectors, direction cosines, dot/cross products, and distance between points/lines. Expect straightforward calculation questions, not proof-based. Speed matters — these questions are solvable in 2-3 minutes each if you know formulas.

Position Vectors and Direction Cosines

A position vector locates a point in 3D space using coordinates (x, y, z). Direction cosines are the cosines of angles a line makes with the three coordinate axes. They always satisfy l² + m² + n² = 1, where l, m, n are direction cosines. Think of direction cosines as 'direction fingerprints' — every line has a unique set. Direction ratios are proportional to direction cosines but don't have to sum to 1.

  • Position vector of point (x, y, z) is r = xi + yj + zk
  • Direction cosines l, m, n satisfy l² + m² + n² = 1 always
  • Direction ratios a:b:c convert to direction cosines by dividing by sqrt(a² + b² + c²)
  • If a line passes through (x₁, y₁, z₁) with direction ratios a:b:c, parametric form is (x₁+at, y₁+bt, z₁+ct)
  • Magnitude of vector v = sqrt(x² + y² + z²)
Key formulas
Direction Cosines from Ratios
l = a/sqrt(a²+b²+c²), m = b/sqrt(a²+b²+c²), n = c/sqrt(a²+b²+c²)
When: Converting direction ratios to direction cosines
Magnitude of Vector
|v| = sqrt(x² + y² + z²)
When: Finding length of any vector
Worked examples

Direction ratios 1:2:2. Find direction cosines. Answer: l=1/3, m=2/3, n=2/3 (divide each by sqrt(1+4+4)=3)

Point A(1,2,3), point B(4,6,11). Find position vector AB. Answer: AB = 3i + 4j + 8k, magnitude = sqrt(9+16+64) = sqrt(89)

Dot Product (Scalar Product)

Dot product of two vectors gives a scalar (single number). It measures how much two vectors point in the same direction. If vectors are perpendicular, dot product is zero — this is the key test for perpendicularity. Dot product is commutative: a·b = b·a. Geometrically, a·b = |a||b|cos(θ), where θ is angle between them.

  • a·b = a₁b₁ + a₂b₂ + a₃b₃ (component form)
  • a·b = |a||b|cos(θ) (geometric form)
  • If a·b = 0, vectors are perpendicular
  • Dot product is always a number, never a vector
  • Used to find angle between lines, check perpendicularity, find projections
Key formulas
Dot Product Component Form
a·b = a₁b₁ + a₂b₂ + a₃b₃
When: When vectors are given in component form (i, j, k)
Angle Between Vectors
cos(θ) = (a·b) / (|a||b|)
When: Finding angle between two vectors or two lines
Projection of a on b
proj_b(a) = (a·b / |b|²) × b
When: Finding component of one vector along another
Worked examples

a = 2i + 3j + k, b = i - j + 2k. Find a·b. Answer: 2(1) + 3(-1) + 1(2) = 2 - 3 + 2 = 1

Check if 2i + j - k and i - 2j - 4k are perpendicular. Dot product = 2(1) + 1(-2) + (-1)(-4) = 2 - 2 + 4 = 4 ≠ 0, so NOT perpendicular

Cross Product (Vector Product)

Cross product of two vectors gives another vector perpendicular to both. It's anti-commutative: a × b = -(b × a). The magnitude |a × b| = |a||b|sin(θ) equals the area of parallelogram formed by the two vectors. Cross product is zero if vectors are parallel. Use determinant method with i, j, k for calculation.

  • a × b is perpendicular to both a and b
  • a × b = -(b × a) (order matters — anti-commutative)
  • If a × b = 0, vectors are parallel
  • |a × b| = |a||b|sin(θ) = area of parallelogram
  • Calculated using 3×3 determinant with i, j, k as first row
Key formulas
Cross Product Determinant
a × b = |i j k| = i(a₂b₃ - a₃b₂) - j(a₁b₃ - a₃b₁) + k(a₁b₂ - a₂b₁)
When: Computing cross product from components
Magnitude of Cross Product
|a × b| = |a||b|sin(θ)
When: Finding area of parallelogram or triangle
Worked example

a = i + 2j + 3k, b = 2i - j + k. Find a × b. Answer: i(2·1 - 3·(-1)) - j(1·1 - 3·2) + k(1·(-1) - 2·2) = i(5) - j(-5) + k(-5) = 5i + 5j - 5k

Distance Between Points and Lines

Distance between two points is straightforward: use the distance formula. Distance from a point to a line requires the perpendicular from the point to the line. Distance between two skew lines (non-parallel, non-intersecting lines in 3D) uses a special formula involving cross product. These are high-frequency exam questions.

  • Distance between P(x₁,y₁,z₁) and Q(x₂,y₂,z₂): sqrt((x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²)
  • Distance from point to line uses perpendicular foot concept
  • Distance between skew lines: |((r₂-r₁)·(d₁×d₂))| / |d₁×d₂|, where r₁, r₂ are position vectors and d₁, d₂ are direction vectors
  • If lines intersect, distance = 0
  • If lines are parallel, use point-to-line formula
Key formulas
Distance Between Two Points
d = sqrt((x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²)
When: Finding distance between any two points in 3D
Distance from Point to Line
d = |(P-A) × d| / |d|, where A is point on line, d is direction vector, P is given point
When: Finding perpendicular distance from a point to a line
Distance Between Skew Lines
d = |((r₂-r₁)·(d₁×d₂))| / |d₁×d₂|
When: Finding distance between two non-intersecting, non-parallel lines
Worked examples

Distance between (1,2,3) and (4,6,11): sqrt((4-1)² + (6-2)² + (11-3)²) = sqrt(9+16+64) = sqrt(89)

Line passes through (1,0,0) with direction 1:1:1. Distance from (2,2,2) to this line: Use cross product method. Vector from (1,0,0) to (2,2,2) is (1,2,2). Direction d = (1,1,1). Cross product (1,2,2)×(1,1,1) = (0,1,-1), magnitude sqrt(2). Distance = sqrt(2)/sqrt(3) = sqrt(2/3)

Equation of Line and Plane

A line in 3D can be written in vector form r = a + λb (where a is position vector of a point, b is direction vector, λ is parameter) or Cartesian form (x-x₁)/l = (y-y₁)/m = (z-z₁)/n. A plane is defined by a normal vector and a point, or by equation ax + by + cz + d = 0. Exam questions often ask for equation of line through two points or plane through three points.

  • Vector form of line: r = a + λb
  • Cartesian form of line: (x-x₁)/l = (y-y₁)/m = (z-z₁)/n
  • Equation of plane: a(x-x₁) + b(y-y₁) + c(z-z₁) = 0, where (a,b,c) is normal vector
  • General plane equation: ax + by + cz + d = 0
  • To find line through two points, use their coordinates to get direction ratios
Key formulas
Vector Equation of Line
r = a + λb
When: Expressing line in vector form
Cartesian Equation of Line
(x-x₁)/l = (y-y₁)/m = (z-z₁)/n
When: Expressing line in Cartesian form
Equation of Plane
a(x-x₁) + b(y-y₁) + c(z-z₁) = 0
When: Plane with normal vector (a,b,c) passing through (x₁,y₁,z₁)
⚠ Common mistakes to avoid
  • Confusing direction cosines with direction ratios — direction cosines ALWAYS satisfy l²+m²+n²=1, direction ratios do not. Always normalize direction ratios first.
  • Getting cross product sign wrong — a × b ≠ b × a. The order matters. Many students forget the negative sign in anti-commutativity.
  • Using dot product formula when cross product is needed, or vice versa — dot product gives a scalar (check perpendicularity), cross product gives a vector (check parallelism or find area).
  • Forgetting to take magnitude in distance formulas — distance is always positive. Don't forget the square root or absolute value bars.
  • Mixing up parametric and Cartesian forms — parametric uses parameter t or λ, Cartesian uses equal ratios. Know which form the question asks for.
🧠 Memory aids
  • DOT = scalar, CROSS = vector. Dot product gives a number (perpendicular when zero). Cross product gives a vector (parallel when zero).
  • Direction Cosines: l² + m² + n² = 1 ALWAYS. Think 'Pythagorean constraint' — they're like coordinates on a unit sphere.
  • Cross Product Determinant: First row is i, j, k. Second row is first vector components. Third row is second vector components. Expand like a 3×3 matrix.
  • Distance Between Skew Lines: Remember the formula has cross product in numerator (magnitude) and denominator (normalizer). It's the 'perpendicular sandwich' formula.
🎯 AGNIVEER NAVY exam tips
  • Agniveer Navy typically asks 2-3 questions on vectors. Expect one straightforward dot/cross product calculation, one distance problem, and possibly one direction cosines question.
  • Time pressure is real — these questions should take 2-3 minutes max. Memorize the determinant expansion for cross product; don't derive it during exam.
  • Recent papers show preference for numerical answers (e.g., 'distance is sqrt(89)') over symbolic. Always simplify your final answer.
  • Direction cosines questions often appear as 'find direction cosines of line through two points' — convert to direction ratios first, then normalize.
  • Skew lines and distance between lines are less common but high-value if you know the formula. If you see it, you can solve it quickly and gain an edge.

Sample questions

Q1 · medium · AI-verified
The midpoint of the line segment joining points A(2, 4, −6) and B(−2, 0, 4) is:
  1. (0, 2, −1)
  2. (0, 4, −1)
  3. (2, 2, −1)
  4. (0, 2, 1)
Q2 · easy · AI-verified
What is the magnitude of the vector **a** = 3î + 4ĵ?
  1. 5
  2. 7
  3. 25
  4. √7
Q3 · easy · AI-verified
What is the dot product of vectors **a** = 2î + 3ĵ and **b** = 4î + 5ĵ?
  1. 23
  2. 26
  3. 15
  4. 20
Q4 · hard · AI-verified
A line makes angles α, β, γ with x, y, z axes respectively. If cos α = 1/2 and cos β = 1/√2, then cos γ = ?
  1. 1/2
  2. √3/2
  3. 0
  4. 1/√2
Q5 · medium · AI-verified
If vectors a⃗ = 3î + 2ĵ − k̂ and b⃗ = î − ĵ + 2k̂, then a⃗ · b⃗ is:
  1. −1
  2. 1
  3. 3
  4. −3
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