 # Scalar product for SQA Higher Maths 1. Introduction to scalar product
2. Scalar product and Cartesian coordinates
3. Angle between two 2D vectors
4. Angle between two 3D vectors

The scalar product of two vectors a and b, denoted a · b, is defined as | a | | b | cos θ. The scalar product of two vectors a = xi + yj + zk and b = pi + qj + wk, denoted a · b, is defined as xp + yq + zw. If two vectors a and b intersect at an angle of θ, then cosθ = a · b/| a | | b | where a · b is the scalar product of vectors a and b. If two vectors a and b intersect at an angle of θ, then cosθ = a · b/|a| |b| where a · b is the scalar product of vectors a and b. # 1

What is the angle between vectors OA and OB where the coordinates of point A is (3, 1) and B is (2, −2) with the origin O? Give your answer in radians to 2 decimal places..

OA · OB = 3 × 2 −2 = 4, | OA | = √(3² + 1²) = √10, | OB | = √( 2² + (−2)² ) = √8. So cosθ = 4/( √10 × √8 ) which leads to θ = 1.11 radians. # 2

There is an unit cube ABCD−A₁B₁C₁D₁, vector AB = a, AD = b, AA₁ = c, find the angle θ between the straight lines A₁B and AC.

| a = | b | = | c | = 1, a · b = b · c = c · a = 0 since cos90º = 0. A₁B = a − c, AC = a + b, so A₁B·AC = (a − c)·(a + b) = | a |² + a · b − a · c − b · c = 1. On the other hand, | A₁B | = | AC | = √2, so cosθ = 1/(√2·√2) = 1/√2, which leads to θ = 60º. # 3

Three points A, B, C have coordinates (1, 5), (3, 7), (2, 3), find the angle between vectors AB and AC. Give your answer in radians to 2 decimal places.

AB = (3 − 1)i + (7 − 5)j = 2i + 2j, AC = (2 − 1)i + (3 − 5)j = i − 2j, AB · AC = 2 − 4 = −2. So cosθ = AB · AC/| AB | | AC | = −2/(√8 × √5) = −√10/10, which leads to θ = 1.89 radians. # 4

| a | = 1, | b | = 3 and a + b = (√3, 1). What is the angle between a + b and a − b?

cosθ = ( (a + b) · (a − b) )/( | (a + b) | · | (a − b) | ) = −1/2. The θ = 2/3π. # 5

Vectors a = 6i + j − 4k, b = 3i − 2j, calculate a · b.

a · b = (6 × 3) −2 = 16. End of page