Vectors
Vectors
Vectors appear in two and three dimensions: magnitude, unit vectors, the scalar product and its use to find angles and prove perpendicularity, plus vector equations of lines. The exam rewards clear component work and correct interpretation of a negative scalar product.
Core rules
- If a = xi + yj + zk then |a| = √(x² + y² + z²); the unit vector is a/|a|.
- Scalar product: a·b = |a||b|cosθ = x₁x₂ + y₁y₂ + z₁z₂, so cosθ = (a·b)/(|a||b|).
- a·b = 0 means a and b are perpendicular (for non-zero vectors); a = kb means they are parallel.
- The position vector of the point dividing AB in the ratio λ:μ is (μa + λb)/(λ + μ); the midpoint is (a + b)/2.
- The line through a with direction d is r = a + td.
Worked example
Given a = 3i − 2j + k and b = i + 4j − 2k, find the angle between them. a·b = (3)(1) + (−2)(4) + (1)(−2) = 3 − 8 − 2 = −7. |a| = √(9 + 4 + 1) = √14 and |b| = √(1 + 16 + 4) = √21. So cosθ = −7/√294 ≈ −0.4083, giving θ ≈ 114.1°. The negative product signals an obtuse angle immediately - a useful check before computing.
Traps
- Reporting the obtuse angle when the acute angle between two lines was requested (use |a·b|).
- Confusing the direction vector of a line with a point on it.
- Forgetting the square root when finding magnitude.
- Treating i, j, k as variables rather than mutually perpendicular unit vectors.