Vectors: Class 12 Mathematics Notes, Short Questions and Numericals

Scalars and vectors in two and three dimensions, unit vectors i, j, k, magnitude, addition, scalar product, vector product, angle between vectors, projection, work done, and area of a parallelogram and a triangle.

Key points

A scalar has magnitude only, such as mass, time and temperature. A vector has both magnitude and direction, such as displacement, velocity and force. A vector is shown by a directed line segment, and written a or AB with an arrow. Its magnitude is written |a|.

In space we use three mutually perpendicular unit vectors i, j and k along the x, y and z axes. Any vector can be written a = a₁i + a₂j + a₃k, or (a₁, a₂, a₃). The magnitude is |a| = √(a₁² + a₂² + a₃²). The position vector of a point P(x, y, z) is OP = xi + yj + zk. The vector from A to B is AB = OB − OA.

A unit vector has magnitude 1. The unit vector in the direction of a is â = a/|a|. The zero vector 0 has magnitude 0 and no direction. Two vectors are equal if they have the same magnitude and direction, which means equal components.

Operations on vectors:

Addition: (a₁, a₂, a₃) + (b₁, b₂, b₃) = (a₁ + b₁, a₂ + b₂, a₃ + b₃). Geometrically use the triangle or parallelogram law.
Subtraction: a − b = a + (−b).
Multiplication by a scalar k: k a = (k a₁, k a₂, k a₃). It changes the magnitude by |k|, and reverses direction if k < 0.
Two vectors are parallel if one is a scalar multiple of the other.

The scalar (dot) product of two vectors is a · b = |a||b| cos θ, where θ is the angle between them (0° ≤ θ ≤ 180°). The result is a scalar. In components, a · b = a₁b₁ + a₂b₂ + a₃b₃. Also a · a = |a|², and i · i = j · j = k · k = 1 while i · j = j · k = k · i = 0.

Properties of the dot product:

Commutative: a · b = b · a.
Distributive: a · (b + c) = a · b + a · c.
Two non-zero vectors are perpendicular if and only if a · b = 0.
The angle between two vectors is found from cos θ = (a · b)/(|a||b|).

Work done by a constant force F over a displacement d is W = F · d, measured in joules when F is in newtons and d in metres. The projection (component) of a on b is (a · b)/|b|.

The vector (cross) product of two vectors is a × b = |a||b| sin θ n̂, where n̂ is the unit vector perpendicular to both a and b, in the direction given by the right hand rule. The result is a vector. Also a × a = 0, and i × j = k, j × k = i, k × i = j.

In components, a × b is found from the determinant with the rows i j k, a₁ a₂ a₃ and b₁ b₂ b₃. This gives a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k. The cross product is not commutative: b × a = −(a × b). It is distributive over addition. Two non-zero vectors are parallel if and only if a × b = 0.

Area from the cross product: the area of the parallelogram with adjacent sides a and b is |a × b|. The area of the triangle with these sides is ½ |a × b|. The moment (torque) of a force F acting at position r is r × F.

Scalar triple product: a · (b × c) gives the volume of the parallelepiped on the three vectors, and it is zero if the three vectors lie in one plane. Check any cross product by testing that it gives zero when dotted with each of the original vectors.

Study tip: Set out components in a column or a clear bracket and compute each one on its own line. Always check a cross product by dotting it with the two original vectors, which should give zero. For work and angle questions, name the formula first, then substitute. Give units for work (joules) and area (square units).

Definitions and flashcards

Say the meaning first, then open the card to check.

Scalar

A quantity with magnitude only.

Vector

A quantity with both magnitude and direction.

Unit vector

A vector of magnitude 1.

Position vector

The vector from the origin to a given point.

Scalar product

The product a · b = |a||b| cos θ, which is a scalar.

Vector product

The product a × b, a vector perpendicular to both a and b.

Projection

The component of one vector along another, (a · b)/|b|.

Parallelogram law

The rule that the sum of two vectors is the diagonal of the parallelogram they form.

Formula sheet

Name	Formula	What it means
Magnitude	\|a\| = √(a₁² + a₂² + a₃²)	Length of the vector.
Unit vector	â = a/\|a\|	Direction of a with magnitude 1.
Dot product	a · b = a₁b₁ + a₂b₂ + a₃b₃ = \|a\|\|b\| cos θ	A scalar.
Angle	cos θ = (a · b)/(\|a\|\|b\|)	Angle between two vectors.
Cross product	a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k	A vector perpendicular to both.
Parallelogram area	A = \|a × b\|	Triangle area is half of this.
Work	W = F · d	In joules.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Distinguish between a scalar and a vector with two examples each. (2 marks)

A scalar has magnitude only, for example mass and time. A vector has magnitude and direction, for example velocity and force.

Find the unit vector in the direction of a = 3i + 4j. (2 marks)

|a| = √(9 + 16) = 5. The unit vector is a/|a| = (3i + 4j)/5 = 0.6i + 0.8j.

Show that the vectors 2i + j − k and i − j + k are perpendicular. (2 marks)

Their dot product is 2(1) + 1(−1) + (−1)(1) = 2 − 1 − 1 = 0. Since the dot product of two non-zero vectors is zero, they are perpendicular.

State the differences between the scalar product and the vector product. (5 marks)

The scalar product a · b = |a||b| cos θ gives a scalar, and the vector product a × b = |a||b| sin θ n̂ gives a vector. The scalar product is commutative, so a · b = b · a, but the vector product is not, since b × a = −(a × b). The scalar product is zero for perpendicular vectors, but the vector product is zero for parallel vectors. The scalar product is used for angle and work done, and the vector product for area and torque.

Write the formula for the area of a triangle with sides a and b taken from one vertex. (3 marks)

The triangle is half of the parallelogram formed by a and b. The area of the parallelogram is |a × b|, so the area of the triangle is ½ |a × b| square units.

Numericals with worked steps

Find a · b for a = 2i + 3j − k and b = i − 2j + 4k.

a · b = a₁b₁ + a₂b₂ + a₃b₃.
= (2)(1) + (3)(−2) + (−1)(4).
= 2 − 6 − 4 = −8.

Answer: −8

Find a × b for a = 2i + 3j − k and b = i − 2j + 4k.

i component: a₂b₃ − a₃b₂ = (3)(4) − (−1)(−2) = 12 − 2 = 10.
j component: −(a₁b₃ − a₃b₁) = −[(2)(4) − (−1)(1)] = −[8 + 1] = −9.
k component: a₁b₂ − a₂b₁ = (2)(−2) − (3)(1) = −4 − 3 = −7.
Check: dot with a gives 20 − 27 + 7 = 0, and with b gives 10 + 18 − 28 = 0.

Answer: 10i − 9j − 7k

Find the angle between a = i + j and b = j + k.

a · b = (1)(0) + (1)(1) + (0)(1) = 1.
|a| = √(1 + 1) = √2 and |b| = √(1 + 1) = √2.
cos θ = 1/(√2 × √2) = 1/2.
θ = 60°.

Answer: 60°

Find the area of the parallelogram with adjacent sides a = i + 2j and b = j + 3k.

a = (1, 2, 0) and b = (0, 1, 3).
i component: (2)(3) − (0)(1) = 6.
j component: −[(1)(3) − (0)(0)] = −3.
k component: (1)(1) − (2)(0) = 1.
a × b = 6i − 3j + k, and |a × b| = √(36 + 9 + 1) = √46 ≈ 6.78.

Answer: √46 ≈ 6.78 square units

A force F = 2i + j + 3k newtons moves a body from A(1, 1, 1) to B(3, 2, 0), with distances in metres. Find the work done.

Displacement d = AB = OB − OA = (3 − 1, 2 − 1, 0 − 1) = (2, 1, −1).
W = F · d = (2)(2) + (1)(1) + (3)(−1).
= 4 + 1 − 3 = 2.

Answer: 2 J

Common mistakes

Writing the dot product as a vector, or the cross product as a number. The dot product is a scalar and the cross product is a vector.
Forgetting the minus sign for the j component when expanding the determinant for a × b.
Assuming a × b = b × a. They are equal in size but opposite in direction.
Using the position vector of B minus A in the wrong order. AB = OB − OA, not OA − OB.
Dividing by |a| only, and not by |a||b|, when finding the angle between two vectors.

Practise this chapter

Practice questions and timed test Printable worksheet