3.7 Vectors
- Syllabus
- 9709–2028–2029
- Topic
- 3.7
- Level
- A2
A vector has size and direction; in 2D or 3D write components as a column or i,j,k combination. The magnitude of (a,b,c) is √(a²+b²+c²).
Keep vectors distinct from points, which describe position relative to an origin. A unit vector has magnitude 1 and preserves direction.
The vector (3,4) has magnitude 5; its unit vector is (3/5,4/5).
The coordinates of a point and the components of a displacement are not interchangeable unless the origin and direction are specified.
Add and subtract vectors componentwise, multiply by a scalar to change magnitude or reverse direction, and use the dot product a·b=|a||b|cosθ to test perpendicularity.
A zero vector has no direction; scalar multiples are parallel. Keep dimensions and coordinate order consistent in every operation.
(2,−1)+(−3,4)=(−1,3); vectors are perpendicular when their dot product is zero.
The dot product is a scalar, not a vector, and a negative scalar reverses direction rather than merely reducing length.
The position vector of P is OP, the directed vector from the origin O to P. The displacement from A to B is OB−OA, and the midpoint is the average of the two position vectors.
Choose one origin, preserve direction in subtraction, and translate geometric statements into vector equations such as a point dividing AB in a given ratio.
If OA=(2,1) and OB=(8,4), then AB=(6,3) and the midpoint has position (5,2.5).
AB is not the same as BA; reversing the order changes the sign of the displacement.
A line through point a in direction d is r=a+λd. Different parameter values move along the same line, and a non-zero scalar multiple of d gives the same direction.
Use the position vector and direction vector separately; compare lines by solving component equations for a common point and compatible parameters.
r=(1,2)+λ(3,−1) passes through (1,2) and has direction (3,−1).
A point on a line is not itself a direction vector, and changing the parameter origin changes a but not the geometric line.
Two vector lines are parallel when their direction vectors are scalar multiples. They are identical only if they also share a point; otherwise they are distinct parallel lines.
Compare direction components first, then test a point from one line in the other. In 3D, skew lines are neither parallel nor intersecting.
r=(1,0)+λ(2,3) and r=(4,5)+μ(−4,−6) are parallel because directions are proportional; checking the point decides whether they coincide.
Equal gradients in a 2D graph are not enough to prove coincident lines; intercepts matter.
For vectors a and b, a·b=|a||b|cosθ and in components a₁b₁+a₂b₂(+a₃b₃). A zero scalar product means perpendicular non-zero vectors.
Use the dot product to find an angle, a projection or a perpendicular condition. Check that the angle is the requested principal angle and that neither vector is zero.
(1,2)·(2,−1)=0, so the vectors are perpendicular; their magnitudes do not need to be equal.
The scalar product is not a vector and a negative value means an obtuse angle, not an impossible result.