Vectors

Maths Pure 3 · Vectors. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A vector in three dimensions is written with i, j and k parts. Its magnitude is its length, and a unit vector has length 1. The vector from A to B is b − a.

A straight line is described by r = a + tb. Here a is the position vector of one point on the line and b is the direction of the line. Two lines in space can be parallel, can intersect, or can be skew, which means not parallel and never meeting.

The scalar product of two vectors is a number. It gives the angle between two directions, and it is zero when they are perpendicular. This fact is used to find the foot of a perpendicular from a point to a line.

Rules to remember

Magnitude: |xi + yj + zk| = √(x² + y² + z²); unit vector = vector ÷ its magnitude
Vector AB = b − a; the midpoint of AB has position vector ½(a + b)
Line: r = a + tb, where a is a point on the line and b is its direction; the line through A and B has direction b − a
Lines are parallel if one direction vector is a multiple of the other; to test for intersection, equate two components, solve for s and t, then check the third component; if the check fails and the lines are not parallel, they are skew
Scalar product: a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ
a·b = 0 means the vectors are perpendicular; for the angle between two lines use only their direction vectors
Foot of the perpendicular N from P to a line: write N as a general point on the line, then solve PN·b = 0 for t; the distance is |PN|

Common mistake

To find the angle between two lines, students use the position vectors of points on the lines. Use the two direction vectors only (the vectors multiplied by s and t). If cos θ comes out negative, the acute angle is 180° − θ.

Worked example

Find the perpendicular distance from the point P(7, 2, 3) to the line r = i + 2j + t(2i + j + 2k).

A general point N on the line is (1 + 2t, 2 + t, 2t), so PN = (2t − 6)i + tj + (2t − 3)k.
PN is perpendicular to the direction 2i + j + 2k: 2(2t − 6) + t + 2(2t − 3) = 0, so 9t − 18 = 0 and t = 2.
N is (5, 4, 4) and PN = −2i + 2j + k.
|PN| = √(4 + 4 + 1) = √9 = 3.

Answer: The distance is 3 units

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Chapter 7: Vectors

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