Vector | NEB Class 11 Mathematics Notes

Chapter Overview

Vector Analysis is a fundamental chapter spanning Mathematics, Physics, Engineering, and Computer Graphics. Unlike ordinary numbers (scalars), vectors possess both magnitude and direction. This chapter introduces vector representations, collinear, coplanar, linear dependence, section formulas, and geometric applications. Mastering this will guarantee solid marks in your Class 11 NEB Mathematics exam.

What this chapter covers

Scalars and VectorsPosition Vectors & Null VectorsMagnitude and Unit VectorsVector Addition & Triangle LawSection Formula (Internal & External)Collinear Points & VectorsCoplanar VectorsLinear Dependence & Independence

Learning Objectives

1

Differentiate between scalar and vector quantities and represent them geometrically.

2

Calculate the magnitude of vectors in 2D and 3D space, and determine unit vectors.

3

Use the 'End − Start' rule to find position vectors and vectors between two points.

4

Apply the section formula to find points dividing a line segment internally or externally.

5

Test for the collinear of three points using scalar multiples.

6

Examine the coplanar and linear dependence of vectors using the Determinant method.

7

Solve geometric proofs (e.g., parallelogram properties, triangle inequality) using vector methods.

The Core Concept: Redundancy vs Uniqueness

Key Insight

The most important concept in Vector Analysis is Linear Dependence vs Linear Independence.

If vectors are Linearly Dependent, one vector can be 'built' by mixing the others. It means they are redundant (e.g., they lie flat on the same plane — Coplanar).

If they are Linearly Independent, every vector brings a unique direction. None can be formed from the others (e.g., spanning 3D space).

Exam Trick: In 3D space, if the determinant of three vectors equals 0, they are stuck in a flat plane (Coplanar) and are Linearly Dependent!

Key Definitions

Scalar Quantity

A physical quantity having only magnitude. Examples: Mass, Length, Area, Temperature.

Vector Quantity

A quantity having both magnitude and direction, represented by a directed line segment $\vec{AB}$. Examples: Velocity, Force, Displacement.

Position Vector

A vector joining the origin $O$ to a specific point $A$. Written as $\vec{OA}$.

Null (Zero) Vector

A vector whose initial and terminal points coincide. Its magnitude is zero ($|\vec{0}| = 0$) and it has an arbitrary direction.

Unit Vector

A vector whose magnitude is exactly 1. It defines pure direction and is denoted by $\hat{a}$. Formula: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.

Collinear Vectors

Vectors that lie on the same line or parallel lines. Condition: $\vec{b} = k\vec{a}$ (one is a scalar multiple of the other).

Coplanar Vectors

Three or more vectors lying entirely within the same plane, or parallel to the same plane.

Linear Combination

An expression formed by multiplying vectors by scalars and adding them: $\alpha_1\vec{a}_1 + \alpha_2\vec{a}_2 + \cdots + \alpha_n\vec{a}_n$.

Linear Dependence vs Independence

Understanding whether a set of vectors is dependent or independent is crucial for geometric proofs and 3D geometry.

Linearly Dependent

$$\alpha_1\vec{a} + \alpha_2\vec{b} + \alpha_3\vec{c} = 0$$

Equation holds true even if some scalars ($\alpha$) are not zero.

One vector can be expressed as a linear combination of the others.

Geometrically: The vectors are Coplanar (flattened into a plane).

Determinant of their coefficients equals 0.

VS

Linearly Independent

$$\alpha_1\vec{a} + \alpha_2\vec{b} + \alpha_3\vec{c} = 0$$

Equation is ONLY true if every single scalar is exactly zero ($\alpha_1=\alpha_2=\alpha_3=0$).

No vector can be built from the others; each provides a unique direction.

Geometrically: Three vectors span an entire 3D volume (Non-coplanar).

Determinant of their coefficients does not equal 0.

Memory Rule: Dependent = Determinant Zero = Flat (Coplanar).

Formula Sheet & Core Derivations

Magnitude of a Vector (3D)

|\vec{a}| = \sqrt{x^2 + y^2 + z^2}

Calculates the physical length of the vector using the 3D Pythagorean theorem.

1A vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ represents a movement of $x$ along the X-axis, $y$ along the Y-axis, and $z$ along the Z-axis.

2By applying the Pythagorean theorem in the XY plane, the base diagonal is $\sqrt{x^2 + y^2}$.

3Applying it again with the Z-axis gives the total length: $|\vec{a}| = \sqrt{(\sqrt{x^2 + y^2})^2 + z^2} = \sqrt{x^2 + y^2 + z^2}$.

Unit Vector

\hat{a} = \frac{\vec{a}}{|\vec{a}|}

Scales a vector down so its length is exactly 1, preserving only its direction.

1Any vector $\vec{a}$ can be thought of as its magnitude $|\vec{a}|$ multiplied by its pure direction $\hat{a}$.

2Therefore, $\vec{a} = |\vec{a}| \times \hat{a}$.

3Dividing both sides by the scalar magnitude $|\vec{a}|$ isolates the unit vector $\hat{a}$.

Section Formula (Internal Division)

\vec{OP} = \frac{m\vec{b} + n\vec{a}}{m + n}

Finds the position vector of point P dividing line segment AB in ratio m:n.

1Let $P$ divide the line segment $AB$ internally such that $\frac{AP}{PB} = \frac{m}{n}$. Let $\vec{OA} = \vec{a}$, $\vec{OB} = \vec{b}$, and $\vec{OP} = \vec{p}$.

2Cross multiplying the ratio gives: $n\vec{AP} = m\vec{PB}$.

3Using the 'End − Start' rule for vectors: $\vec{AP} = \vec{OP} - \vec{OA}$ and $\vec{PB} = \vec{OB} - \vec{OP}$.

4Substitute these in: $n(\vec{OP} - \vec{a}) = m(\vec{b} - \vec{OP})$.

5Expand and group terms with $\vec{OP}$: $n\vec{OP} + m\vec{OP} = m\vec{b} + n\vec{a}$.

6Factor out $\vec{OP}$: $(m+n)\vec{OP} = m\vec{b} + n\vec{a}$.

7Divide by $(m+n)$ to get the final formula: $\vec{OP} = \frac{m\vec{b} + n\vec{a}}{m+n}$.

coplanar via Determinant

\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0

If the determinant of the components of three 3D vectors is 0, they are coplanar.

1Assume three vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent (meaning they are coplanar).

2This implies we can write one as a combination of the others: $\vec{c} = x\vec{a} + y\vec{b}$.

3Writing this in component form: $c_1\hat{i} + c_2\hat{j} + c_3\hat{k} = x(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) + y(b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$.

4Equating components gives a system of linear equations: $a_1x + b_1y - c_1 = 0$, $a_2x + b_2y - c_2 = 0$, $a_3x + b_3y - c_3 = 0$.

5For this system to have a consistent solution for $x$ and $y$, the determinant of their coefficients must be zero. Hence the determinant test.

Checking Vector Relationships

When asked to prove collinear, coplanar, or dependence, follow this flowchart.

Are you given POINTS (A, B, C) or VECTORS (a, b, c)?

Yes

POINTS -> Find vectors first using End − Start (e.g., AB = OB - OA). Then proceed ↓

No

VECTORS -> Proceed directly ↓

Are you checking for collinear (2 directions)?

Yes

Check if one vector is a scalar multiple of another (b = k * a). If Yes -> Collinear.

No

Proceed to next question ↓

Are you checking for coplanar / DEPENDENCE (3 directions)?

Yes

Calculate the 3x3 determinant of their components. Det = 0 -> Coplanar/Dependent. Det ≠ 0 -> Independent.

No

Use standard addition/section formulas.

The 'End − Start' Rule

Trick: To find a vector between any two points, subtract the position vector of the starting point from the ending point.

Steps to Remember:

To find vector $\vec{AB}$, identify the End (B) and Start (A).
Subtract: $\vec{AB} = \vec{OB} - \vec{OA}$.
Example: If $A(1, 2)$ and $B(4, 6)$, then $\vec{OA} = (1, 2)$ and $\vec{OB} = (4, 6)$.
$\vec{AB} = (4, 6) - (1, 2) = (4-1, 6-2) = (3, 4)$.

Mnemonic: AB means 'A to B'. Subtract A from B. End minus Start.

Collinear = Scaling Ratio

Trick: Two vectors are collinear if their respective $x, y, z$ components form the exact same ratio.

Steps to Remember:

Let $\vec{a} = (x_1, y_1, z_1)$ and $\vec{b} = (x_2, y_2, z_2)$.
Check the ratios: $\frac{x_2}{x_1} = \frac{y_2}{y_1} = \frac{z_2}{z_1} = k$.
If all three ratios are equal to the same constant $k$, the vectors are collinear.
Example: $\vec{a} = (1, 3, -5)$ and $\vec{b} = (2, 6, -10)$. Ratios: $2/1 = 2$, $6/3 = 2$, $-10/-5 = 2$. Constant ratio $k=2$, so they are collinear.

Mnemonic: Collinear vectors are just resized versions of each other.

Common Mistakes (Lose Marks Here!)

Avoid these extremely common student errors spotted in NEB answer sheets.

Confusing Points with Vectors: When asked if points A, B, C are collinear, you cannot just check scalar multiples of A and B directly. You MUST first construct the vectors $\vec{AB}$ and $\vec{AC}$, then check if $\vec{AC} = k\vec{AB}$.
Wrong Unit Vector Formula: Writing $\hat{a} = \frac{|a|}{a}$ is a fatal error. A scalar cannot be divided by a vector. It is always Vector ÷ Magnitude: $\frac{\vec{a}}{|\vec{a}|}$.
Section Formula Flip: Students often write $\frac{m\vec{a} + n\vec{b}}{m+n}$. This is WRONG. It must be a cross-multiplication: the ratio $m$ (on the left) multiplies vector $\vec{b}$ (on the right). Correct: $\frac{m\vec{b} + n\vec{a}}{m+n}$.
Determinant Sign Errors: When expanding a 3x3 determinant manually, remember the alternating signs for the top row: ( + , − , + ). Forgetting the minus sign on the middle term ruins the calculation.

NEB Exam Traps for 2083

NEB Exam Warning

Board examiners intentionally design questions to test if you truly understand these distinctions.

Coplanar vs Collinear: 'Collinear' applies to points on a line. 'Coplanar' applies to vectors in a plane. All collinear vectors are coplanar, but NOT all coplanar vectors are collinear. Use the correct test!
The 'Examine Dependence' Trap: If an NEB question asks you to 'Examine if the vectors are linearly dependent', showing that the determinant is 0 is enough for full marks. Do not waste time trying to solve simultaneous equations for $x, y, z$ unless specifically asked to 'express as a linear combination'.
Null Vector Trap: Any set of vectors that contains the null vector $\vec{0}$ is ALWAYS linearly dependent. This frequently appears as a tricky 1-mark MCQ.

Triangle Law of Vector Addition

The Triangle Law states that if two vectors are represented as two sides of a triangle in sequence (tip-to-tail), their sum (resultant) is represented by the third side taken in the opposite order (from the start of the first to the end of the second).

AB + BC = AC. Notice how tracing the path A → B → C gives the exact same total displacement as going straight from A → C.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Find position vector of...	Use OA, OB relative to origin O.
Find unit vector in direction of...	Compute magnitude first, then use Vector ÷ Magnitude.
Show points A, B, C are collinear	Find vectors AB and AC. Prove AC = k * AB.
Examine if vectors are coplanar	Set up 3x3 Determinant. If det = 0, they are coplanar.
Linearly dependent / independent	Use the determinant method (Dependent = 0, Independent ≠ 0).
Find the resultant vector	Add the given vectors algebraically component by component.
Divide a line segment in ratio m:n	Use the Section Formula: (m b + n a) / (m + n).

Solved Examples (Easy → Hard)

1

Find the value of $m$ if the magnitude of the vector $(m, -6)$ is 10.

Easy

Solution

The vector is given by $\vec{a} = m\hat{i} - 6\hat{j}$.
We know the formula for magnitude: $|\vec{a}| = \sqrt{x^2 + y^2}$.

Given $|\vec{a}| = 10$:
$$10 = \sqrt{m^2 + (-6)^2}$$
Squaring both sides to remove the root:
$$100 = m^2 + 36$$
$$m^2 = 100 - 36 = 64$$
$$m = \pm 8$$

∴ m = 8 or m = -8

2

Find a unit vector parallel to the resultant of vectors $\vec{a} = (2, 4, 3)$ and $\vec{b} = (1, -3, -1)$.

Medium

Solution

Step 1 — Find the resultant vector $\vec{r}$:
The resultant is the sum of the vectors.
$\vec{r} = \vec{a} + \vec{b} = (2+1, 4-3, 3-1) = (3, 1, 2)$

Step 2 — Find the magnitude of $\vec{r}$:
$|\vec{r}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

Step 3 — Find the unit vector $\hat{r}$:
$\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{1}{\sqrt{14}}(3, 1, 2)$

∴ Unit vector = (3/√14, 1/√14, 2/√14)

3

Examine if the vectors $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, $\vec{b} = 3\hat{i} - 5\hat{j} + 2\hat{k}$, and $\vec{c} = 4\hat{i} - 5\hat{j} + \hat{k}$ are linearly dependent.

Hard

Solution

To test for linear dependence of three 3D vectors, we set up the determinant of their scalar components.

$$D = \begin{vmatrix} 2 & -3 & 1 \\ 3 & -5 & 2 \\ 4 & -5 & 1 \end{vmatrix}$$

Expand along the first row (remember alternating signs +, -, +):
$$D = 2[(-5)(1) - (2)(-5)] - (-3)[(3)(1) - (2)(4)] + 1[(3)(-5) - (-5)(4)]$$
$$D = 2[-5 + 10] + 3[3 - 8] + 1[-15 + 20]$$
$$D = 2(5) + 3(-5) + 1(5)$$
$$D = 10 - 15 + 5 = 0$$

Since the determinant is exactly 0, the vectors are stuck in the same plane (coplanar).

∴ Since D = 0, the vectors are Linearly Dependent.

MCQ Practice (10 Questions)

Mixed Level — MCQ

1

If $\vec{a} = k\hat{a}$, then what is the value of k?

Correct!

Incorrect. Correct: |a|

Solution

Every vector consists of a magnitude (scalar) and a direction (unit vector).
$\vec{a} = \text{Magnitude} \times \text{Direction} = |\vec{a}| \hat{a}$.
Comparing this to $\vec{a} = k\hat{a}$, we see $k = |\vec{a}|$.

∴ |a|

2

What is the magnitude of a unit vector?

Correct!

Incorrect. Correct: 1

Solution

A unit vector, denoted by $\hat{n}$, is defined mathematically as a vector of length 1.
$$|\hat{n}| = 1$$

∴ 1

3

Which of the following is the algebraic condition for two non-zero vectors a and b to be collinear?

Correct!

Incorrect. Correct: b = k a

Solution

Collinear vectors are parallel. Parallel vectors maintain the same proportional direction, meaning one vector can be multiplied by a scalar $k$ to perfectly match the other.
Condition: $\vec{b} = k\vec{a}$.

∴ b = k a

4

If the determinant formed by the coefficients of three vectors is not equal to zero, the vectors are:

Correct!

Incorrect. Correct: Linearly Independent

Solution

The determinant of three 3D vectors calculates the volume of the parallelepiped they form. If the volume is non-zero (Det ≠ 0), the vectors don't lie flat on a single plane, hence they are Linearly Independent.

∴ Linearly Independent

5

Any set of vectors that contains a null (zero) vector is always:

Correct!

Incorrect. Correct: Linearly Dependent

Solution

Consider the set $\{\vec{0}, \vec{a}, \vec{b}\}$.
We can form the linear combination: $1(\vec{0}) + 0(\vec{a}) + 0(\vec{b}) = \vec{0}$.
Because one of the scalars (the 1) is NOT zero, the set is Linearly Dependent by definition.

∴ Linearly Dependent

6

For what value of m are the vectors (1, -2, 1), (2, 1, -1), and (7, -m, 1) linearly dependent?

Correct!

Incorrect. Correct: 4

Solution

Set up determinant and equate to zero:
$\begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -1 \\ 7 & -m & 1 \end{vmatrix} = 0$
Expand along row 1:
$1(1(1) - (-1)(-m)) - (-2)(2(1) - (-1)(7)) + 1(2(-m) - 1(7)) = 0$
$1(1 - m) + 2(2 + 7) + 1(-2m - 7) = 0$
$1 - m + 18 - 2m - 7 = 0$
$12 - 3m = 0 \Rightarrow 3m = 12 \Rightarrow m = 4$.

∴ 4

7

If $\vec{AC} = -\vec{AB}$, what is the relationship between points A, B, and C?

Correct!

Incorrect. Correct: A is the midpoint of BC

Solution

Given: $\vec{AC} = -\vec{AB}$.
We know that reversing the direction of a vector flips its sign: $-\vec{AB} = \vec{BA}$.
So, $\vec{AC} = \vec{BA}$.
This means the displacement from B to A is identical in length and direction to the displacement from A to C. They lie on a straight line, and A is exactly in the middle. Therefore, A is the midpoint of BC.

∴ A is the midpoint of BC

8

If vector r = x(vector a) + y(vector b), what can we conclude about vectors r, a, and b?

Correct!

Incorrect. Correct: They are coplanar

Solution

The equation $\vec{r} = x\vec{a} + y\vec{b}$ states that $\vec{r}$ is a linear combination of $\vec{a}$ and $\vec{b}$.
Geometrically, any linear combination of two vectors results in a vector that lies perfectly within the flat plane generated by those two vectors. Therefore, $\vec{r}$, $\vec{a}$, and $\vec{b}$ are coplanar.

∴ They are coplanar

9

For $\vec{i} + 3\hat{j} - 5\hat{k}$ and $2\hat{i} + m\hat{j} - 10\hat{k}$ to be collinear, the value of m must be:

Correct!

Incorrect. Correct: 6

Solution

For two vectors to be collinear, their respective components must be proportional:
$\frac{x_2}{x_1} = \frac{y_2}{y_1} = \frac{z_2}{z_1}$
Plugging in our values:
$\frac{2}{1} = \frac{m}{3} = \frac{-10}{-5}$
$2 = \frac{m}{3} = 2$
$\frac{m}{3} = 2 \Rightarrow m = 6$.

∴ 6

10

Three non-coplanar vectors are ALWAYS:

Correct!

Incorrect. Correct: Linearly Independent

Solution

Coplanar vectors are Linearly Dependent. Therefore, the exact opposite—non-coplanar vectors—must be Linearly Independent. They span an entire 3-dimensional space.

∴ Linearly Independent

Short Questions (2 Marks Each)

1

Define linearly dependent vectors.

[2 marks]

Solution

A set of vectors is linearly dependent if at least one vector in the set can be expressed as a linear combination of the others.
Mathematically, for vectors $\vec{a}_1, \vec{a}_2, \dots, \vec{a}_n$, the equation
$\alpha_1\vec{a}_1 + \alpha_2\vec{a}_2 + \dots + \alpha_n\vec{a}_n = 0$
holds true where at least one of the scalars $\alpha_i$ is not zero.

∴ A set where one vector is a combination of others.

2

Find the value of k if the vectors $\vec{a} = (2, 4)$ and $\vec{b} = (1, k)$ are collinear.

[2 marks]

Solution

For two vectors to be collinear, their components must be strictly proportional.
Therefore, the ratio of their x-components must equal the ratio of their y-components:
$$\frac{x_a}{x_b} = \frac{y_a}{y_b}$$
$$\frac{2}{1} = \frac{4}{k}$$
Cross-multiplying gives:
$$2k = 4 \Rightarrow k = 2$$

∴ k = 2

3

How do you test if three vectors are linearly independent?

[2 marks]

Solution

To test three given vectors in 3D space for linear independence, arrange their scalar components into a $3 \times 3$ determinant.
Evaluate the determinant:
1. If Det ≠ 0, the vectors are Linearly Independent.
2. If Det = 0, the vectors are Linearly Dependent.

∴ Check if their 3x3 determinant is non-zero.

4

Find the vector $\vec{AB}$ and its modulus if point A is (1, 2, -3) and point B is (5, -4, 7).

[2 marks]

Solution

Using the 'End − Start' rule:
$$\vec{AB} = \vec{OB} - \vec{OA} = (5, -4, 7) - (1, 2, -3)$$
$$\vec{AB} = (5-1, -4-2, 7-(-3)) = (4, -6, 10)$$

Now, find the modulus (magnitude):
$$|\vec{AB}| = \sqrt{4^2 + (-6)^2 + 10^2}$$
$$|\vec{AB}| = \sqrt{16 + 36 + 100} = \sqrt{152} = 2\sqrt{38}$$

∴ AB = (4, -6, 10); Modulus = 2√38

5

Are the standard basis vectors $\hat{i}=(1,0,0)$, $\hat{j}=(0,1,0)$, and $\hat{k}=(0,0,1)$ linearly dependent or independent?

[2 marks]

Solution

Set up the determinant of their components:
$$D = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$$
Expanding along the first row gives:
$$D = 1(1 \times 1 - 0) = 1$$
Since $D = 1 \neq 0$, the vectors are linearly independent. They span the entire 3D space.

∴ Linearly Independent

6

Define coplanar vectors. What happens to their determinant?

[2 marks]

Solution

Coplanar vectors are three or more vectors that lie on the exact same plane (or are parallel to the same plane).
Because they are restricted to 2 dimensions mathematically, one vector can always be expressed as a linear combination of the other two.
Consequently, the determinant formed by their components is exactly zero.

∴ Vectors in same plane. Determinant = 0.

7

If points A, B, and C are collinear in that order, what geometric relation connects $\vec{AB}$, $\vec{BC}$, and $\vec{AC}$?

[2 marks]

Solution

By the Triangle Law of Vector Addition, for any three points, $\vec{AC} = \vec{AB} + \vec{BC}$.
Furthermore, because they are collinear, the magnitude of the total vector is the simple sum of the individual lengths:
$$|\vec{AC}| = |\vec{AB}| + |\vec{BC}|$$

∴ AC = AB + BC

8

Find the unit vector in the direction of $\vec{a} = 3\hat{i} - 4\hat{j}$.

[2 marks]

Solution

Step 1 — Find magnitude:
$$|\vec{a}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Step 2 — Apply unit vector formula:
$$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{3\hat{i} - 4\hat{j}}{5} = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}$$

∴ 3/5 i - 4/5 j

Long Questions (3 / 4 Marks Each)

1

Prove that the points $A(1, 2, 4)$, $B(2, 5, -1)$, and $C(3, 8, -6)$ are collinear.

[3 marks]

Solution

Step 1 — Find vector $\vec{AB}$:
$\vec{AB} = \vec{OB} - \vec{OA} = (2-1, 5-2, -1-4) = (1, 3, -5)$

Step 2 — Find vector $\vec{AC}$:
$\vec{AC} = \vec{OC} - \vec{OA} = (3-1, 8-2, -6-4) = (2, 6, -10)$

Step 3 — Establish the scalar relationship:
Notice that $(2, 6, -10)$ is exactly 2 times $(1, 3, -5)$.
Therefore, $\vec{AC} = 2\vec{AB}$.

Because $\vec{AC}$ is a scalar multiple of $\vec{AB}$, the two vectors are parallel. Furthermore, since they share the common point A, they must lie on the exact same straight line. Hence, A, B, and C are collinear.

∴ Proved via AC = 2AB.

2

Prove that for any parallelogram ABCD, $\vec{AC} + \vec{BD} = 2\vec{BC}$.

[3 marks]

Solution

Let $ABCD$ be a parallelogram.

Step 1 — Express diagonals using vector addition:
By the Triangle Law in $\triangle ABC$: $\vec{AC} = \vec{AB} + \vec{BC}$. (Equation 1)
By the Triangle Law in $\triangle BCD$: $\vec{BD} = \vec{BC} + \vec{CD}$. (Equation 2)

Step 2 — Add the two equations:
$$\vec{AC} + \vec{BD} = (\vec{AB} + \vec{BC}) + (\vec{BC} + \vec{CD})$$
$$\vec{AC} + \vec{BD} = \vec{AB} + 2\vec{BC} + \vec{CD}$$

Step 3 — Apply parallelogram properties:
In a parallelogram, opposite sides are equal and parallel, but $\vec{AB}$ and $\vec{CD}$ point in opposite directions as defined by the vertex order. Therefore, $\vec{AB} = -\vec{CD}$.

Substitute this into our sum:
$$\vec{AC} + \vec{BD} = -\vec{CD} + 2\vec{BC} + \vec{CD}$$
The $\vec{CD}$ terms cancel out, leaving:
$$\vec{AC} + \vec{BD} = 2\vec{BC} \quad \blacksquare$$

∴ Proved using Triangle Law.

3

Show that vectors $5\vec{a} + 6\vec{b} + 7\vec{c}$, $7\vec{a} - 8\vec{b} + 9\vec{c}$, and $3\vec{a} + 20\vec{b} + 5\vec{c}$ are coplanar.

[4 marks]

Solution

Let the vectors be $\vec{V}_1$, $\vec{V}_2$, and $\vec{V}_3$. If they are coplanar, one can be expressed as a linear combination of the other two: $\vec{V}_3 = x\vec{V}_1 + y\vec{V}_2$.

$$3\vec{a} + 20\vec{b} + 5\vec{c} = x(5\vec{a} + 6\vec{b} + 7\vec{c}) + y(7\vec{a} - 8\vec{b} + 9\vec{c})$$

Equating the coefficients of $\vec{a}$, $\vec{b}$, and $\vec{c}$ gives a system of 3 equations:
1) $5x + 7y = 3$
2) $6x - 8y = 20$
3) $7x + 9y = 5$

Solve Eq 1 and Eq 2:
Multiply Eq 1 by 6: $30x + 42y = 18$
Multiply Eq 2 by 5: $30x - 40y = 100$
Subtract: $82y = -82 \Rightarrow y = -1$.
Substitute $y = -1$ into Eq 1: $5x - 7 = 3 \Rightarrow 5x = 10 \Rightarrow x = 2$.

Check with Eq 3:
$7(2) + 9(-1) = 14 - 9 = 5$. This perfectly matches Eq 3!

Since $x=2$ and $y=-1$ satisfy all equations, $\vec{V}_3 = 2\vec{V}_1 - \vec{V}_2$. The vectors are coplanar.

∴ Proved via linear combination.

4

Prove the Triangle Inequality: $|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$.

[4 marks]

Solution

Geometric Proof:
Consider a triangle $OAB$ formed by vectors.
Let $\vec{OA} = \vec{a}$ and $\vec{AB} = \vec{b}$.
By the Triangle Law of Vector Addition, the closing side is $\vec{OB} = \vec{a} + \vec{b}$.

In any geometrical triangle, the sum of the lengths of any two sides is always greater than the length of the third side. Therefore:
$$|OB| < |OA| + |AB|$$
$$|\vec{a} + \vec{b}| < |\vec{a}| + |\vec{b}|$$

When do they become strictly equal? If vectors $\vec{a}$ and $\vec{b}$ are collinear and point in the same direction, the triangle 'flattens' into a straight line. Then, their lengths add up directly:
$$|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}|$$

Combining both the triangle and flat-line cases gives the overall inequality:
$$|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}| \quad \blacksquare$$

∴ Proved geometrically.

5

Find $m$ so that the vectors $2\hat{i} - \hat{j} + \hat{k}$, $\hat{i} + 2\hat{j} - 3\hat{k}$, and $3\hat{i} + 2m\hat{j} + 5\hat{k}$ are coplanar.

[3 marks]

Solution

For three vectors to be coplanar, the determinant of their components must be equal to zero.

$$D = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & 2m & 5 \end{vmatrix} = 0$$

Expand along the first row:
$$2[ (2)(5) - (-3)(2m) ] - (-1)[ (1)(5) - (-3)(3) ] + 1[ (1)(2m) - (2)(3) ] = 0$$
$$2[ 10 + 6m ] + 1[ 5 + 9 ] + 1[ 2m - 6 ] = 0$$
$$20 + 12m + 14 + 2m - 6 = 0$$

Combine like terms:
$$14m + 28 = 0$$
$$14m = -28$$
$$m = -2$$

∴ m = -2

6

Express $\vec{r} = (4, 7)$ as a linear combination of $\vec{a} = (5, -4)$ and $\vec{b} = (-2, 5)$.

[3 marks]

Solution

To express $\vec{r}$ as a linear combination of $\vec{a}$ and $\vec{b}$, we set up the equation:
$$\vec{r} = x\vec{a} + y\vec{b}$$
$$(4, 7) = x(5, -4) + y(-2, 5)$$

Equating the $x$ and $y$ components gives two simultaneous equations:
1) $5x - 2y = 4$
2) $-4x + 5y = 7$

Solve the system:
Multiply Eq 1 by 4: $20x - 8y = 16$
Multiply Eq 2 by 5: $-20x + 25y = 35$
Add both equations: $17y = 51 \Rightarrow y = 3$.

Substitute $y = 3$ into Eq 1:
$5x - 2(3) = 4 \Rightarrow 5x - 6 = 4 \Rightarrow 5x = 10 \Rightarrow x = 2$.

Thus, the linear combination is: $\vec{r} = 2\vec{a} + 3\vec{b}$.

∴ r = 2a + 3b

7

Derive the Section Formula for internal division of a line segment.

[4 marks]

Solution

Let $P$ be a point that divides the line segment $AB$ internally in the ratio $m:n$.
Let $O$ be the origin. The position vectors are $\vec{OA} = \vec{a}$, $\vec{OB} = \vec{b}$, and $\vec{OP} = \vec{r}$.

Because $P$ divides $AB$ in $m:n$, we know:
$$\frac{AP}{PB} = \frac{m}{n} \Rightarrow n\vec{AP} = m\vec{PB}$$

Using the position vectors (End - Start):
$\vec{AP} = \vec{OP} - \vec{OA} = \vec{r} - \vec{a}$
$\vec{PB} = \vec{OB} - \vec{OP} = \vec{b} - \vec{r}$

Substitute these into our ratio equation:
$$n(\vec{r} - \vec{a}) = m(\vec{b} - \vec{r})$$
$$n\vec{r} - n\vec{a} = m\vec{b} - m\vec{r}$$

Move all terms containing $\vec{r}$ to the left side:
$$m\vec{r} + n\vec{r} = m\vec{b} + n\vec{a}$$
$$(m+n)\vec{r} = m\vec{b} + n\vec{a}$$

Divide by $(m+n)$ to isolate $\vec{r}$:
$$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n} \quad \blacksquare$$

∴ Derived successfully.

8

Prove that any set of vectors containing a null vector is linearly dependent.

[3 marks]

Solution

Consider a set of vectors $\{ \vec{0}, \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \}$.

By the definition of linear dependence, the set is dependent if there exists a linear combination:
$$c_0\vec{0} + c_1\vec{v}_1 + c_2\vec{v}_2 + \dots + c_n\vec{v}_n = 0$$
where at least one coefficient is not zero.

Let us choose $c_0 = 1$ (which is non-zero) and set all other coefficients ($c_1, c_2, \dots, c_n$) to exactly $0$.

Substitute these in:
$$1(\vec{0}) + 0(\vec{v}_1) + 0(\vec{v}_2) + \dots = \vec{0} + \vec{0} + \vec{0} = \vec{0}$$

The equation balances perfectly to 0. Since we achieved this sum of zero using a non-zero scalar ($c_0 = 1$), the set strictly satisfies the condition for being linearly dependent.

∴ Proved using definition.

9

In a regular hexagon ABCDEF, prove that the resultant of $\vec{AB}$, $\vec{AC}$, $\vec{AD}$, $\vec{AE}$, and $\vec{AF}$ is $6\vec{AO}$, where O is the center.

[4 marks]

Solution

Let $O$ be the center of regular hexagon $ABCDEF$.
In a regular hexagon, opposite vertices form diagonals that pass through $O$. Therefore, $\vec{AD} = 2\vec{AO}$.

We need to find the sum: $S = \vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF}$.
Rearrange the terms symmetrically:
$$S = (\vec{AB} + \vec{AF}) + (\vec{AC} + \vec{AE}) + \vec{AD}$$

By properties of the hexagon, $ABOF$ forms a parallelogram. Thus, the diagonal $\vec{AO} = \vec{AB} + \vec{AF}$.
Similarly, $ACDE$ forms a larger parallelogram structure. By geometry, $\vec{AC} + \vec{AE} = \vec{AD} = 2\vec{AO}$.

Substitute these back into $S$:
$$S = (\vec{AO}) + (2\vec{AO}) + (2\vec{AO})$$
Note: Wait, $\vec{AD} = 2\vec{AO}$.
So $S = \vec{AO} + 2\vec{AO} + 2\vec{AO} = 5\vec{AO}$. Wait! If the question says $6\vec{AO}$, let's recheck $AC+AE$.

Actually, $\vec{AC} = \vec{AB} + \vec{BC}$. And $\vec{AE} = \vec{AF} + \vec{FE}$.
Sum $\vec{AC} + \vec{AE} = (\vec{AB} + \vec{AF}) + (\vec{BC} + \vec{FE})$.
Since $\vec{BC} = \vec{AO}$ and $\vec{FE} = \vec{AO}$, their sum is $2\vec{AO}$.
Also $\vec{AB} + \vec{AF} = \vec{AO}$.
Thus $\vec{AC} + \vec{AE} = \vec{AO} + 2\vec{AO} = 3\vec{AO}$.

Now plug back into $S$:
$$S = (\vec{AB} + \vec{AF}) + (\vec{AC} + \vec{AE}) + \vec{AD}$$
$$S = (\vec{AO}) + (3\vec{AO}) + (2\vec{AO}) = 6\vec{AO} \quad \blacksquare$$

∴ Proved: Sum = 6 AO

10

Find the coordinates of point B if vector $\vec{AB} = 2\hat{i} + 3\hat{j} - 4\hat{k}$ and point A is $(1, 2, 2)$.

[3 marks]

Solution

We know the End-Start rule: $\vec{AB} = \vec{OB} - \vec{OA}$.

Given:
$\vec{AB} = (2, 3, -4)$
$\vec{OA} = (1, 2, 2)$

We need to find $\vec{OB}$. Rearranging the formula gives:
$$\vec{OB} = \vec{AB} + \vec{OA}$$

Add the corresponding components:
$$\vec{OB} = (2, 3, -4) + (1, 2, 2)$$
$$\vec{OB} = (2+1, 3+2, -4+2) = (3, 5, -2)$$

Thus, the coordinates of point B are $(3, 5, -2)$.

∴ B = (3, 5, -2)

Past NEB Board Questions

1

Find λ if $3\hat{i} + \hat{j} - \hat{k}$ and $\lambda\hat{i} - 4\hat{j} + 4\hat{k}$ are collinear.

NEB 2080 [2 marks]

Solution

For two vectors to be collinear, the ratio of their scalar components must be equal:
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$
$$\frac{3}{\lambda} = \frac{1}{-4} = \frac{-1}{4}$$

Taking the first two parts of the equation:
$$\frac{3}{\lambda} = \frac{-1}{4}$$
Cross-multiplying gives:
$$-\lambda = 12 \Rightarrow \lambda = -12$$

∴ λ = -12

2

Prove that the quadrilateral formed by joining the midpoints of adjacent sides of any quadrilateral is a parallelogram.

NEB 2079 [4 marks]

Solution

Let $ABCD$ be a quadrilateral. Let $P, Q, R, S$ be the midpoints of $AB, BC, CD, DA$ respectively.
Let the origin be $O$. Position vectors of midpoints are:
$\vec{p} = \frac{\vec{a}+\vec{b}}{2}$, $\vec{q} = \frac{\vec{b}+\vec{c}}{2}$, $\vec{r} = \frac{\vec{c}+\vec{d}}{2}$, $\vec{s} = \frac{\vec{a}+\vec{d}}{2}$.

Find vector $\vec{PQ}$:
$\vec{PQ} = \vec{q} - \vec{p} = \frac{\vec{b}+\vec{c}}{2} - \frac{\vec{a}+\vec{b}}{2} = \frac{\vec{c} - \vec{a}}{2} = \frac{1}{2}\vec{AC}$.

Find vector $\vec{SR}$:
$\vec{SR} = \vec{r} - \vec{s} = \frac{\vec{c}+\vec{d}}{2} - \frac{\vec{a}+\vec{d}}{2} = \frac{\vec{c} - \vec{a}}{2} = \frac{1}{2}\vec{AC}$.

Since $\vec{PQ} = \vec{SR}$, sides $PQ$ and $SR$ are both parallel and equal in length. Hence $PQRS$ is a parallelogram.

∴ Proved via midpoint vectors.

3

Show points A(1, 2, 3), B(3, 4, 7), and C(-3, -2, -5) are collinear.

NEB 2078 [3 marks]

Solution

Find $\vec{AB}$ and $\vec{AC}$.
$\vec{AB} = (3-1, 4-2, 7-3) = (2, 2, 4)$
$\vec{AC} = (-3-1, -2-2, -5-3) = (-4, -4, -8)$

Notice that $\vec{AC} = -2(2, 2, 4)$.
Therefore, $\vec{AC} = -2\vec{AB}$.

Since $\vec{AC}$ is a scalar multiple of $\vec{AB}$, they are parallel and share point A. Thus, they are collinear.

∴ Proved AC = -2 AB.

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes under exam conditions to prepare for the NEB finals.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

If vector a and vector b are collinear, then vector b = ?

Correct!

Incorrect. Correct: k * a

Solution

Two vectors are collinear if they possess the same line of support. This requires their directions to be the same or exactly opposite, differing only by a scale factor $k$.

∴ k * a

2

What does a determinant of exactly 0 mean for three vectors?

Correct!

Incorrect. Correct: They are Coplanar

Solution

A determinant equal to 0 mathematically proves the vectors are linearly dependent, meaning they are Coplanar.

∴ They are Coplanar

2 Section B — Short Questions (2 Marks Each)

1

Find the modulus of vector v = (-2, 3, -4).

[2 marks]

Solution

$|v| = \sqrt{(-2)^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.

∴ √29

2

Write down the Section Formula for internal division.

[2 marks]

Solution

For a point dividing segment AB internally in ratio m:n, the position vector is:
$\vec{OP} = \frac{m\vec{b} + n\vec{a}}{m + n}$.

∴ (m b + n a) / (m + n)

3 Section C — Long Questions (4 Marks Each)

1

Show that the vectors $a-3b+5c$, $a-2b+3c$, and $-2a+3b-4c$ are coplanar.

[4 marks]

Solution

Let $V_1, V_2, V_3$ be the vectors.
If coplanar, $\det = 0$. Using the coefficients of $a, b, c$:
$D = \begin{vmatrix} 1 & -3 & 5 \\ 1 & -2 & 3 \\ -2 & 3 & -4 \end{vmatrix}$
Expand: $1(8-9) - (-3)(-4+6) + 5(3-4) = 1(-1) + 3(2) + 5(-1) = -1 + 6 - 5 = 0$.
Since determinant is 0, they are coplanar.

∴ Proved via determinant = 0.

2

Test the collinear of points P(2, 1, -1), Q(3, -2, 1), and R(1, 4, -3).

[4 marks]

Solution

Find $\vec{PQ} = (3-2, -2-1, 1-(-1)) = (1, -3, 2)$.
Find $\vec{PR} = (1-2, 4-1, -3-(-1)) = (-1, 3, -2)$.
Notice that $\vec{PR} = -(1, -3, 2) = -\vec{PQ}$.
Since $\vec{PR}$ is a scalar multiple of $\vec{PQ}$ ($k = -1$), they are collinear.

∴ Proved PR = -PQ.

One-Page Revision Cheat Sheet

1

Vector = Magnitude (size) + Direction.

2

Modulus (Length): $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$.

3

Unit Vector: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$. (Always divide the vector by its magnitude, not the reverse!)

4

End - Start Rule: $\vec{AB} = \vec{OB} - \vec{OA}$.

5

Collinear Vectors: Points on a line or parallel. Condition: $\vec{b} = k\vec{a}$. Check if their $x,y,z$ ratios are equal.

6

Coplanar Vectors: Vectors flattened onto a single plane. Condition: Determinant of their $3 \times 3$ matrix equals exactly 0.

7

Linearly Dependent: A redundant system where one vector is a sum of others. (Det = 0).

8

Linearly Independent: A unique system spanning 3D volume. (Det ≠ 0).

9

Section Formula (Internal): $\vec{OP} = \frac{m\vec{b} + n\vec{a}}{m + n}$. (Remember to cross-multiply ratios and vectors!).

10

Triangle Law: Head-to-tail addition. $\vec{AB} + \vec{BC} = \vec{AC}$.