Vectors in two and three dimensions flashcards

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Vector
A quantity with both magnitude and direction, written in i, j, k components or as a column vector.
Unit vector
A vector of magnitude 1. The unit vector in the direction of a is a ÷ |a|.
Position vector of a point A
The vector OA from the origin O to A; its components are the coordinates of A.
Parallel vectors
Vectors that are scalar multiples of each other: b = ka for some non-zero scalar k.
Skew lines
Lines in three dimensions that are not parallel and do not intersect.
Foot of the perpendicular from P to a line
The point F on the line for which PF is perpendicular to the line; it is the point of the line nearest to P.
Magnitude of xi + yj + zk
√(x² + y² + z²) (No — learn it)
Vector from A to B in terms of position vectors
AB = b − a (end minus start) (No — learn it)
Distance between (x₁, y₁, z₁) and (x₂, y₂, z₂)
√((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²) (No — learn it)
Vector equation of a line
r = a + tb: a is the position vector of a point on the line, b a direction vector, t a scalar parameter. (No — learn it)
Scalar product of a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k
a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ (No — learn it)
Position vector of the point dividing AB in the ratio λ : μ
a + (λ/(λ + μ))(b − a) = (μa + λb)/(λ + μ) (No — derive it)
Area of triangle ABC using vectors
½ |AB| |AC| sin θ, where θ is the angle between AB and AC (from the scalar product). (No — learn it)
Magnitude of a 3D vector as a diagram
Two right-angled triangles: the floor diagonal √(x² + y²), then the vertical z, giving √(x² + y² + z²).
The angle θ in a·b = |a||b| cos θ
The angle between the two vectors drawn tail to tail (both starting from the same point).
Two lines: the three cases in 3D
Intersecting (meet at one point), parallel (multiple directions, no common point) or skew (neither).
Image of P in a line
F is the foot of the perpendicular and the midpoint of PP′, so OP′ = 2OF − OP. (No — derive it)
Shortest distance from a point to a line
The length of the perpendicular PF from the point to the line.
How do you test whether a point lies on r = a + tb?
Find t from one component, then check the same t gives the other two components.
How do you show three points A, B, C are collinear?
Show AC (or BC) is a scalar multiple of AB; with the common point A (or B) the points are on one line.
How do you show two lines are skew?
Show the directions are not parallel, then that the parameters from two component equations fail the third equation.
Why is a·b = 0 for perpendicular vectors?
a·b = |a||b| cos 90° and cos 90° = 0.
Which vectors give the angle between two lines?
Their direction vectors only; if the scalar product is negative, the acute angle is 180° − θ.
How do you find the foot F of the perpendicular from P to r = a + tb?
Write F in terms of t, form PF, solve PF·b = 0 for t, then substitute back.
Unit vector in the direction of 2i − 3j + 6k
|v| = √(4 + 9 + 36) = 7, so the unit vector is (1/7)(2i − 3j + 6k).
Distance between A(3, 3, 1) and B(7, −1, 8)
AB = 4i − 4j + 7k, so AB = √(16 + 16 + 49) = 9.
Are 2i + j − 2k and 3i + 4j + 5k perpendicular?
Scalar product 6 + 4 − 10 = 0, so yes.
Angle between i + k and j + k
Scalar product 1; magnitudes √2 and √2; cos θ = 1/2, so θ = 60°.
Does (7, 1, 4) lie on r = (i + 4j − 2k) + t(2i − j + 2k)?
x: 1 + 2t = 7 gives t = 3; y: 4 − 3 = 1 ✓; z: −2 + 6 = 4 ✓. Yes, at t = 3.
Values of p for which |pi + 2j − 2k| = √17
p² + 4 + 4 = 17, so p² = 9 and p = 3 or p = −3.
AB written as a − b
Wrong way round: AB = b − a (end minus start).
Using the same parameter for both lines
Give the lines different parameters (λ and μ); the lines can meet at different parameter values.
Stopping after solving two equations for λ and μ
Always substitute into the third equation: it decides between intersecting and skew.
Angle between lines from position vectors
Use the direction vectors, not the position vectors of points on the lines.
Angle at vertex A of a triangle
Use AB and AC (both from A); BA with AC gives 180° minus the angle.
Exact area of a triangle
Find sin θ from cos θ using sin²θ = 1 − cos²θ, or use ½ × base × perpendicular height.

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