ZIMSEC A Level · Pure Mathematics · Form 6
Basics
Vectors in 3D: a = xi + yj + zk. Magnitude |a| = √(x² + y² + z²). A unit vector is a/|a|. Position vectors: AB = b − a.
Scalar (dot) product
a · b = |a||b| cos θ = x₁x₂ + y₁y₂ + z₁z₂. Perpendicular vectors have a · b = 0.
Worked example: Find the angle between a = 2i − j + 2k and b = i + 2j + 2k.
a · b = 2 − 2 + 4 = 4; |a| = 3; |b| = 3
cos θ = 4/9 → θ ≈ 63.6°
Vector equation of a line
r = a + λb, where a is a point on the line and b the direction.
- Two lines may be parallel (direction vectors are multiples), intersect (a common point exists), or be skew (neither).
- To find an intersection, equate components and solve for λ and μ, then check the third component.
Worked example: Find the line through A(1, 0, 2) and B(3, 1, −1).
Direction AB = 2i + j − 3k → r = (i + 2k) + λ(2i + j − 3k)
Foot of a perpendicular
Find the point P on a line so that AP · b = 0; the perpendicular distance is |AP|.
Exam tip: Lines in 3D that do not meet are not necessarily parallel – check for skew lines.
Check yourself
- Find |3i − 4j + 12k|.
- Show that i + 2j − k and 3i − j + k are perpendicular.
- Write the vector equation of the line through (2, −1, 0) parallel to i + 3k.
Show answers
- √(9 + 16 + 144) = 13
- 3 − 2 − 1 = 0, so they are perpendicular.
- r = (2i − j) + λ(i + 3k)
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