StudySquare
Scalar product for SQA Higher Maths
This page covers the following topics:
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.
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