Scalar (dot) product
Work done by a pulling force depends on how much of the force goes along the motion. The dot product measures how much two vectors point together, and gives one number as the answer.
The dot product of a and b is |a| |b| cos θ, where θ is the angle between them. In components, multiply the matching components and add. If a · b = 0 for non-zero vectors, then they are perpendicular. Also a · a = |a|².
Formula: a · b = |a| |b| cos θ = a₁b₁ + a₂b₂ + a₃b₃, cos θ = (a · b) / (|a| |b|)
Rule 1: Multiply and add — Same-axis components only.
- Easy: (2i + 3j) · (i − j) = 2 − 3 = −1.
- Medium: (i + 2j + 3k) · (4i − 5j + 6k) = 4 − 10 + 18 = 12.
- Exam-type: |a| = 3 and |b| = 4 with θ = 60° give a · b = 6.
Rule 2: Angle and perpendicularity — Use cos θ = (a · b)/(|a| |b|).
- Easy: The angle between i + j and j + k is 60°, since cos θ = 1/2.
- Medium: 2i + λj + k is perpendicular to i + 2j − k when 2 + 2λ − 1 = 0, so λ = −½.
- Exam-type: If a ⊥ b with |a| = 3 and |b| = 4, then |a + b| = 5.
The answer of a dot product is a number, not a vector. Perpendicular vectors give 0.
Writing the answer of the dot product as a vector, or using the cross product rule of signs.
Dot gives a number, zero means a right angle.
Practice questions
- 1Find (2i + 3j) · (i − j).
- 2Find (i + 2j + 3k) · (4i − 5j + 6k).
- 3Find the angle between i + j and j + k.
- 4Find λ so that 2i + λj + k is perpendicular to i + 2j − k.
- 5Find a · a for a = 2i − 2j + k.
- 6Find the angle between i and i + j.
- 7Find a · b if |a| = 3, |b| = 4 and θ = 60°.
- 8If |a| = 3, |b| = 4 and a ⊥ b, find |a + b|.
Show answers (try them yourself first)
- 1−1
- 212
- 3cos θ = 1/(√2 × √2) = 1/2, so θ = 60°
- 42 + 2λ − 1 = 0, so λ = −½
- 54 + 4 + 1 = 9
- 6cos θ = 1/√2, so θ = 45°
- 73 × 4 × ½ = 6
- 8√(9 + 16 + 0) = 5
Projection of a vector on a line
A tall pole casts a shadow when the sun shines at an angle. The projection of one vector on another is the length of its shadow on that vector.
The projection of a on b is the number (a · b)/|b|. The vector projection is that number multiplied by the unit vector along b. The projection can be negative if the angle is obtuse.
Formula: Projection of a on b = (a · b)/|b|, vector projection = ((a · b)/|b|²) b
Rule 1: Shadow length — Dot product divided by the length of the base vector.
- Easy: The projection of 2i + 3j on i is 2.
- Medium: The projection of i + j + k on 2i + 3j + 6k is 11/7.
- Exam-type: The projection of 2i − j + k on i + 2j − 2k is −2/3.
Rule 2: Vector projection — Multiply by the unit vector of the base.
- Easy: The vector projection of 3i + 4j on i is 3i.
- Medium: The vector projection of 2i + j on i + j is (3/2)(i + j).
- Exam-type: The projection of i + λj + k on i + 2j + 2k is 3 when λ = 3.
Divide by the magnitude of the vector you project on, not by the magnitude of the first vector.
Dividing by |a| instead of |b|.
Dot divided by the base length.
Practice questions
- 1Find the projection of 2i + 3j on i.
- 2Find the projection of i + j + k on 2i + 3j + 6k.
- 3Find the projection of 3i + 4j on j.
- 4Find the vector projection of 3i + 4j on i.
- 5Find the vector projection of 2i + j on i + j.
- 6Find the projection of 2i − j + k on i + 2j − 2k.
- 7Find the projection of i + j + k on i + j.
- 8Find λ so that the projection of i + λj + k on i + 2j + 2k is 3.
Show answers (try them yourself first)
- 12
- 2(2 + 3 + 6)/7 = 11/7
- 34
- 43i
- 5(3/2)(i + j)
- 6(2 − 2 − 2)/3 = −2/3
- 72/√2 = √2
- 8(1 + 2λ + 2)/3 = 3, so λ = 3