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Vector Algebra — chapter notes

6 sections 4 min readBeta

1. Vectors, magnitude and direction

A vector has magnitude and direction and is drawn as a directed line segment PQ→\overrightarrow{PQ}. For a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}:

  • ∣a⃗∣=a12+a22+a32|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}; the unit vector is a^=a⃗∣a⃗∣\hat{a} = \dfrac{\vec{a}}{|\vec{a}|}.
  • Direction cosines l,m,nl, m, n are the cosines of the angles with the axes: l=a1∣a⃗∣l = \dfrac{a_1}{|\vec{a}|} and so on, with l2+m2+n2=1l^2 + m^2 + n^2 = 1.
  • Direction ratios are any three numbers proportional to l,m,nl, m, n, for example a1,a2,a3a_1, a_2, a_3.
  • A vector of magnitude kk along a⃗\vec{a} is ka^k\hat{a}.
Direction cosine = component ÷ magnitude.
VectorMagnitudeDirection cosines
2i^−6j^+3k^2\hat{i} - 6\hat{j} + 3\hat{k}4+36+9=7\sqrt{4 + 36 + 9} = 727,−67,37\tfrac27, -\tfrac67, \tfrac37
4i^+4j^−7k^4\hat{i} + 4\hat{j} - 7\hat{k}16+16+49=9\sqrt{16 + 16 + 49} = 949,49,−79\tfrac49, \tfrac49, -\tfrac79
i^−2j^+2k^\hat{i} - 2\hat{j} + 2\hat{k}1+4+4=3\sqrt{1 + 4 + 4} = 313,−23,23\tfrac13, -\tfrac23, \tfrac23
Ratios are proportional to the cosines; cosines are fixed.
QuantityFormulaNote
Magnitude∣a⃗∣=a12+a22+a32|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}always ≥0\ge 0
Direction cosinesl=a1∣a⃗∣, m=a2∣a⃗∣, n=a3∣a⃗∣l = \dfrac{a_1}{|\vec{a}|},\ m = \dfrac{a_2}{|\vec{a}|},\ n = \dfrac{a_3}{|\vec{a}|}l2+m2+n2=1l^2 + m^2 + n^2 = 1
Direction ratiosa1,a2,a3a_1, a_2, a_3 (or any multiple)not unique

Formulas

  • Direction cosines
    l2+m2+n2=1l^2 + m^2 + n^2 = 1

Key terms

Unit vector

A vector of magnitude 11.

Direction cosines

Cosines of the angles a vector makes with the xx-, yy- and zz-axes.

2. Types of vectors

  • Zero vector 0⃗\vec{0}: magnitude 00, no definite direction.
  • Unit vector: magnitude 11.
  • Equal vectors: same magnitude and same direction (wherever drawn).
  • Negative of a vector: same magnitude, opposite direction; BA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}.
  • Collinear (parallel) vectors: b⃗=λa⃗\vec{b} = \lambda\vec{a} for a scalar λ\lambda — like when λ>0\lambda > 0, opposite when λ<0\lambda < 0. In components, the ratios b1a1=b2a2=b3a3\dfrac{b_1}{a_1} = \dfrac{b_2}{a_2} = \dfrac{b_3}{a_3} are equal.
Collinear vectors may point the same way or opposite ways.
TypeMeaningExample
Zeromagnitude 00, no definite direction0⃗\vec{0}
Unitmagnitude 1135i^+45j^\tfrac35\hat{i} + \tfrac45\hat{j}
Equalsame magnitude and directionAB→=DC→\overrightarrow{AB} = \overrightarrow{DC} in a parallelogram
Negativesame magnitude, opposite directionBA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}
Collinear (parallel)one is a scalar multiple of the otheri^−2k^\hat{i} - 2\hat{k} and −3i^+6k^-3\hat{i} + 6\hat{k}

Formulas

  • Collinear
    b⃗=λa⃗\vec{b} = \lambda\vec{a}

Key terms

Collinear vectors

Vectors along parallel lines; one is a scalar multiple of the other.

3. Position vectors and operations

The position vector of A(x,y,z)A(x, y, z) is a⃗=xi^+yj^+zk^\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}, and AB→=b⃗−a⃗\overrightarrow{AB} = \vec{b} - \vec{a} (head minus tail).

  • Triangle law: AB→+BC→=AC→\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}; the parallelogram law gives the same sum as a diagonal.
  • In components, add, subtract and scale one direction at a time.
  • The distance between AA and BB is ∣AB→∣|\overrightarrow{AB}|.
$\overrightarrow{AB} = \vec{b} - \vec{a}$.
Work with each of $\hat{i}$, $\hat{j}$, $\hat{k}$ separately.
OperationRule (component-wise)Example with a⃗=i^+2j^−k^\vec{a} = \hat{i} + 2\hat{j} - \hat{k}, b⃗=2i^−j^\vec{b} = 2\hat{i} - \hat{j}
Sumadd components3i^+j^−k^3\hat{i} + \hat{j} - \hat{k}
Differencesubtract components−i^+3j^−k^-\hat{i} + 3\hat{j} - \hat{k}
Scalar multiplemultiply each component3a⃗=3i^+6j^−3k^3\vec{a} = 3\hat{i} + 6\hat{j} - 3\hat{k}
Triangle law: $\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB}$.

Formulas

  • Joining two points
    AB→=b⃗−a⃗\overrightarrow{AB} = \vec{b} - \vec{a}

Key terms

Position vector

The vector from the origin to a point; its components are the coordinates.

4. The section formula

If RR divides PQPQ in the ratio m:nm : n:

  • internally: r⃗=mq⃗+np⃗m+n\vec{r} = \dfrac{m\vec{q} + n\vec{p}}{m + n} (RR between PP and QQ);
  • externally: r⃗=mq⃗−np⃗m−n\vec{r} = \dfrac{m\vec{q} - n\vec{p}}{m - n}, m≠nm \ne n (RR on PQPQ produced).
  • Midpoint: p⃗+q⃗2\dfrac{\vec{p} + \vec{q}}{2}. The centroid of a triangle is a⃗+b⃗+c⃗3\dfrac{\vec{a} + \vec{b} + \vec{c}}{3}.
  • Collinear points: A,B,CA, B, C are collinear when AC→=λ AB→\overrightarrow{AC} = \lambda\,\overrightarrow{AB}; the ratio follows from λ\lambda.
$R$ divides $PQ$ internally in $m : n$ (drawn for $2 : 1$).
The weight $m$ goes with the far end $Q$.
DivisionPosition vector of RRWhere RR lies
Internal, m:nm : nmq⃗+np⃗m+n\dfrac{m\vec{q} + n\vec{p}}{m + n}between PP and QQ
External, m:nm : n (m≠nm \ne n)mq⃗−np⃗m−n\dfrac{m\vec{q} - n\vec{p}}{m - n}on PQPQ produced
Midpointp⃗+q⃗2\dfrac{\vec{p} + \vec{q}}{2}the case m=nm = n, internal
Same weights, different signs and denominators.
Division of PQPQ in 2:32 : 3WorkingPoint
Internal2(6,8,−2)+3(1,−2,3)5=(15,10,5)5\dfrac{2(6, 8, -2) + 3(1, -2, 3)}{5} = \dfrac{(15, 10, 5)}{5}(3,2,1)(3, 2, 1)
External2(6,8,−2)−3(1,−2,3)2−3=(9,22,−13)−1\dfrac{2(6, 8, -2) - 3(1, -2, 3)}{2 - 3} = \dfrac{(9, 22, -13)}{-1}(−9,−22,13)(-9, -22, 13)

Formulas

  • Internal division
    r⃗=mq⃗+np⃗m+n\vec{r} = \dfrac{m\vec{q} + n\vec{p}}{m + n}
  • External division
    r⃗=mq⃗−np⃗m−n\vec{r} = \dfrac{m\vec{q} - n\vec{p}}{m - n}

Key terms

External division

Division of a segment by a point on the segment produced.

5. The scalar (dot) product and projections

  • a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3, a scalar; a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}; a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2.
  • Angle: cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}; non-zero vectors are perpendicular iff a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0.
  • ∣a⃗±b⃗∣2=∣a⃗∣2+∣b⃗∣2±2 a⃗⋅b⃗|\vec{a} \pm \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 \pm 2\,\vec{a} \cdot \vec{b}.
  • Scalar projection of a⃗\vec{a} on b⃗\vec{b}: a⃗⋅b⃗∣b⃗∣\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|} (negative for an obtuse angle). Vector projection: a⃗⋅b⃗∣b⃗∣2 b⃗\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\,\vec{b}.
$OM = |\vec{a}|\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
The answer is a number (a scalar).
PropertyStatement
Definitiona⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta
In componentsa1b1+a2b2+a3b3a_1b_1 + a_2b_2 + a_3b_3
Commutativea⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}
Self producta⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2
Unit vectorsi^⋅i^=1\hat{i} \cdot \hat{i} = 1, i^⋅j^=0\hat{i} \cdot \hat{j} = 0
Perpendiculara⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0 for non-zero a⃗,b⃗\vec{a}, \vec{b}
A negative projection means an obtuse angle.
Projection of a⃗\vec{a} on b⃗\vec{b}FormulaFor a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}, b⃗=i^−2j^+2k^\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}
Scalara⃗⋅b⃗∣b⃗∣\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}−63=−2\dfrac{-6}{3} = -2
Vectora⃗⋅b⃗∣b⃗∣2 b⃗\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\,\vec{b}−23(i^−2j^+2k^)-\dfrac23(\hat{i} - 2\hat{j} + 2\hat{k})

Formulas

  • Angle
    cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

Key terms

Scalar projection

The signed length of the shadow of a⃗\vec{a} on the line of b⃗\vec{b}.

6. The vector (cross) product, areas and collinearity

  • a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}, perpendicular to both (right-hand rule); b⃗×a⃗=−(a⃗×b⃗)\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}).
  • i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j}; i^×i^=0⃗\hat{i} \times \hat{i} = \vec{0}.
  • In components, expand ∣i^j^k^a1a2a3b1b2b3∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}.
  • Unit vectors perpendicular to both: ±a⃗×b⃗∣a⃗×b⃗∣\pm\dfrac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}.
  • Areas: parallelogram ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|; triangle 12∣a⃗×b⃗∣\tfrac12|\vec{a} \times \vec{b}|; with diagonals 12∣d1⃗×d2⃗∣\tfrac12|\vec{d_1} \times \vec{d_2}|.
  • Parallel / collinear: a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}. Also ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2.
Area $= |\vec{a}| \cdot |\vec{b}|\sin\theta = |\vec{a} \times \vec{b}|$; a triangle on the same sides has half of it.
The answer is a vector.
PropertyStatement
Magnitude∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta
Directionperpendicular to both a⃗\vec{a} and b⃗\vec{b} (right-hand rule)
Orderb⃗×a⃗=−(a⃗×b⃗)\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})
Unit vectorsi^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j}
Parallel vectorsa⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}
Halve for a triangle; halve again never.
FigureArea
Parallelogram on adjacent sides a⃗,b⃗\vec{a}, \vec{b}∣a⃗×b⃗∣|\vec{a} \times \vec{b}|
Triangle on adjacent sides a⃗,b⃗\vec{a}, \vec{b}12∣a⃗×b⃗∣\tfrac12|\vec{a} \times \vec{b}|
Triangle ABCABC12∣AB→×AC→∣\tfrac12|\overrightarrow{AB} \times \overrightarrow{AC}|
Parallelogram with diagonals d1⃗,d2⃗\vec{d_1}, \vec{d_2}12∣d1⃗×d2⃗∣\tfrac12|\vec{d_1} \times \vec{d_2}|

Formulas

  • Magnitude
    ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta

Key terms

Right-hand rule

Curl the fingers from a⃗\vec{a} towards b⃗\vec{b}; the thumb points along a⃗×b⃗\vec{a} \times \vec{b}.

Exam tips

How ISC marks are awarded

  • Unit vector, direction cosines (1–2 marks): the magnitude, then the division.
  • Section formula (2–3 marks): the formula with the correct weights and sign, then the point.
  • Angle and projection (2–3 marks): the dot product, the magnitudes, the value (keep its sign).
  • Perpendicular unit vector, area (3 marks): the determinant, the cross product, its magnitude, then the unit vector (both signs) or the area (halved for a triangle).
  • Section D (5 marks): set in parts such as 2 + 3; label each part.

Common mistakes

  • Writing AB→=a⃗−b⃗\overrightarrow{AB} = \vec{a} - \vec{b} (it is head minus tail).
  • Forgetting m−nm - n in the external section formula.
  • Losing the minus sign of the j^\hat{j} term in a cross product.
  • Treating a⃗⋅b⃗\vec{a} \cdot \vec{b} as a vector, or a⃗×b⃗=b⃗×a⃗\vec{a} \times \vec{b} = \vec{b} \times \vec{a}.
  • Dividing by ∣b⃗∣2|\vec{b}|^2 for the scalar projection (that is for the vector projection).