A vector is a quantity with both magnitude and direction − like velocity, force or displacement. In NDA Maths, Vector Algebra is one of the most scoring chapters: the formulas are short, the questions are direct, and a little practice guarantees 2−4 easy marks. This guide builds everything from scratch so you never freeze on a vector question again.
Why Vectors Are Worth Your Time
Every NDA Maths paper carries questions from vectors, and almost all of them are formula-based − no long proofs, no messy algebra. If you memorise about ten formulas and understand what they mean, you can solve these in under a minute each.
Vectors also connect to 3D geometry, physics (force, velocity, work) and even your SSB technical interviews. So the effort you put here pays off across the whole syllabus.
The chapter is small but dense: nearly every line is a usable formula. That is good news for a busy student − there is very little "theory" to read and a lot of direct application. Treat each formula as a tool, learn exactly what it does, and you will recognise the right tool the moment you read a question.
In the NDA paper, vector questions are usually pure plug-and-chug. Keep the dot-product and cross-product formulas at your fingertips and you will rarely lose marks here.
Scalars Versus Vectors
A scalar has only magnitude (size): mass, time, temperature, speed, distance. A vector has magnitude and direction: displacement, velocity, force, acceleration.
We write a vector as a (bold) or with an arrow, a→. Its magnitude (length) is written |a→| or simply a, and is always a non-negative number.
- Zero vector (0→): magnitude 0, no fixed direction.
- Unit vector (â): magnitude exactly 1; it only shows direction.
- Equal vectors: same magnitude and same direction.
- Negative vector (−a→): same magnitude, opposite direction.
Speed is a scalar, velocity is a vector. Distance is a scalar, displacement is a vector. The exam loves to test this difference.
Position Vectors and Component Form
Fix an origin O. The position vector of a point P is OP→, the directed line from O to P. In 3D we use three perpendicular unit vectors î, ĵ, k̂ pointing along the x, y and z axes.
Any vector: a→ = a1î + a2ĵ + a3k̂
Magnitude: |a→| = √(a12 + a22 + a32)
If a point P has coordinates (x, y, z), then OP→ = xî + yĵ + zk̂. The numbers a1, a2, a3 are called the components of the vector.
Vector joining two points
If A = (x1, y1, z1) and B = (x2, y2, z2), then the vector from A to B is found by tip minus tail:
AB→ = (x2−x1)î + (y2−y1)ĵ + (z2−z1)k̂
AB→ means B minus A, not A minus B. Always subtract the starting point from the ending point, or your direction flips.
Adding Vectors and Multiplying by a Scalar
Vectors add component by component:
If a→ = a1î + a2ĵ + a3k̂ and b→ = b1î + b2ĵ + b3k̂, then a→ + b→ = (a1+b1)î + (a2+b2)ĵ + (a3+b3)k̂.
Geometric rules
- Triangle law: place tail of b→ at tip of a→; the sum is the closing side.
- Parallelogram law: if a→ and b→ are adjacent sides, the diagonal from their common point is a→ + b→.
Multiplying by a scalar λ stretches or shrinks: λa→ has magnitude |λ|·|a→|. If λ > 0 the direction stays the same; if λ < 0 it reverses.
Vector addition obeys the same friendly rules as ordinary numbers: it is commutative (a→ + b→ = b→ + a→) and associative ((a→ + b→) + c→ = a→ + (b→ + c→)). Adding the zero vector changes nothing, and adding the negative of a vector brings you back to the origin. These simple properties let you rearrange long vector expressions without fear.
Unit vector along a→: â = a→ ÷ |a→|
This is the single most-used trick in the chapter − it gives direction without changing size.
Section Formula for Vectors
The section formula gives the position vector of a point that divides a line segment in a given ratio − very handy for midpoints and centroids.
If R divides AB in ratio m : n internally, with position vectors a→ and b→:
r→ = (m·b→ + n·a→) ÷ (m + n)
Midpoint (m = n): r→ = (a→ + b→) ÷ 2
For external division, replace the plus with a minus in the numerator and denominator. The centroid of a triangle with vertices a→, b→, c→ is simply (a→ + b→ + c→) ÷ 3.
Notice the ratio numbers swap sides: m goes with b→, n goes with a→. Mixing these up is the most common error in section-formula questions.
The Dot Product (Scalar Product)
The dot product of two vectors gives a number (scalar), not a vector. It measures how much two vectors point in the same direction.
a→ · b→ = |a→| |b→| cosθ (where θ is the angle between them)
In components: a→ · b→ = a1b1 + a2b2 + a3b3
What the dot product tells you
- If a→ · b→ = 0 (and neither is zero), the vectors are perpendicular (θ = 90°).
- Angle between vectors: cosθ = (a→ · b→) ÷ (|a→| |b→|).
- a→ · a→ = |a→|2.
Projection
The projection (scalar component) of a→ on b→ is (a→ · b→) ÷ |b→|. This tells you the "shadow length" of a→ along b→.
Dot product → scalar (a number). Use it for angles and perpendicularity checks.
The Cross Product (Vector Product)
The cross product of two vectors gives a new vector perpendicular to both. Its direction follows the right-hand rule.
|a→ × b→| = |a→| |b→| sinθ
a→ × b→ = (a2b3 − a3b2)î − (a1b3 − a3b1)ĵ + (a1b2 − a2b1)k̂
This is the expansion of the 3×3 determinant with î, ĵ, k̂ in the top row, a-components in the second row and b-components in the third row.
Key uses
- If a→ × b→ = 0→, the vectors are parallel (collinear).
- Area of a parallelogram with sides a→, b→ = |a→ × b→|.
- Area of a triangle with sides a→, b→ = ½ |a→ × b→|.
- A unit vector perpendicular to both = (a→ × b→) ÷ |a→ × b→|.
The cross product is not commutative: a→ × b→ = −(b→ × a→). Swapping the order flips the sign. The dot product, however, is commutative.
Collinear, Perpendicular and Coplanar Vectors
These three conditions appear again and again in the NDA paper, so memorise the tests:
- Collinear (parallel): a→ = λb→ for some scalar λ, OR a→ × b→ = 0→. Components are in equal ratio: a1/b1 = a2/b2 = a3/b3.
- Perpendicular: a→ · b→ = 0.
- Coplanar (three vectors): their scalar triple product a→ · (b→ × c→) = 0.
Scalar triple product [a→ b→ c→] = a→ · (b→ × c→) is a single number equal to the 3×3 determinant of the components. Its absolute value is the volume of the parallelepiped formed by the three vectors.
To check whether three points are collinear, form two vectors from them and test if their cross product is the zero vector. Quick and reliable.
Worked Example: Angle and Area
Let us combine the dot and cross products in one problem, exactly the style NDA likes.
Given a→ = 2î + ĵ + 2k̂ and b→ = î − 2ĵ + 2k̂, find (i) the angle between them and (ii) the area of the triangle formed by them.
Notice how the same two vectors give an angle (via dot) and an area (via cross). Mastering this pair handles most vector questions.
Common Mistakes to Avoid
Most marks lost in vectors come from small, avoidable slips. Watch out for these:
- Treating the dot product as a vector − it is a scalar. The cross product is the vector one.
- Forgetting the middle (ĵ) term is negative when expanding the cross-product determinant.
- Mixing up sin and cos: dot uses cosθ, cross uses sinθ.
- Using AB→ = A − B instead of B − A.
- Forgetting the ½ when finding the area of a triangle (parallelogram has no half).
For a unit vector, students often forget to divide by the magnitude. â = a→ ÷ |a→| − without that division it is not a unit vector.
Previous-Year Style Practice
Here is a question in the exact pattern the NDA examiners use. Try it before reading the answer.
Q. If a→ = î + ĵ + k̂ and b→ = ĵ − k̂, then a vector c→ such that a→ × c→ = b→ and a→ · c→ = 3 makes which check easiest first? Simpler version: find a unit vector perpendicular to both a→ = 2î − ĵ + k̂ and b→ = î + ĵ − k̂.
Answer: Compute a→ × b→. î:(−1)(−1)−(1)(1)=1−1=0; ĵ:−[(2)(−1)−(1)(1)]=−(−3)=3; k̂:(2)(1)−(−1)(1)=2+1=3. So a→ × b→ = 0î + 3ĵ + 3k̂. Its magnitude = √(0+9+9)=√18=3√2. The unit perpendicular vector = (3ĵ + 3k̂) ÷ (3√2) = (ĵ + k̂) ÷ √2.
Whenever a question asks for a vector perpendicular to two given vectors, your first move should always be the cross product.
Quick Revision
Run through this checklist the night before your exam − it covers almost every vector question type.
- Magnitude: |a→| = √(a12+a22+a32); unit vector â = a→ ÷ |a→|.
- Dot: a→·b→ = a1b1+a2b2+a3b3 = |a→||b→|cosθ → angles, perpendicular check.
- Cross: |a→×b→| = |a→||b→|sinθ → perpendicular vector, area.
- Perpendicular ↔ dot = 0; Parallel ↔ cross = 0→.
- Triangle area = ½|a→×b→|; parallelogram area = |a→×b→|.
- Coplanar ↔ scalar triple product = 0; its modulus = parallelepiped volume.
Frequently asked questions
What is the difference between the dot product and the cross product?
The dot product gives a scalar (a number) and uses cosθ; it is best for finding angles and checking perpendicularity. The cross product gives a vector perpendicular to both inputs and uses sinθ; it is best for areas and direction.
How do I check whether two vectors are perpendicular or parallel?
Two non-zero vectors are perpendicular when their dot product equals 0. They are parallel (collinear) when their cross product is the zero vector, or equivalently when their components are in equal ratio.
How do I find a unit vector in the direction of a given vector?
Divide the vector by its own magnitude: â = a→ ÷ |a→|. The result has length 1 and points in the same direction as the original vector.
Why is the middle term of the cross-product determinant negative?
When you expand a 3×3 determinant by the top row, the cofactor signs alternate plus, minus, plus. The ĵ term sits in the middle position, so it carries the minus sign. Forgetting it is a very common exam error.
Is vector algebra an important topic for the NDA exam?
Yes. Vectors appear in every NDA Maths paper and the questions are short and formula-based, making them among the most scoring. With about ten formulas memorised, you can secure these marks quickly and reliably.
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