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3.7 Vectors

Syllabus
9709–2028–2029
Topic
3.7
Level
A2

Learning objectives

3.7.1Standard notations for vectors• use standard notations for vectors, i.e. x yfp, xi + yj, x y z fp, xi + yj + zk, AB, a3.7.2Vector operations• carry out addition and subtraction of vectors and multiplication of a vector by a scalar, and interpret these operations in geometrical terms e.g. 'OABC is a parallelogram' is equivalent to OB OA OC= +. The general form of the ratio theorem is not included, but understanding that the midpoint of AB has position vector OA OB2 1 +_ i is expected.3.7.3Position vectors• calculate the magnitude of a vector, and use unit vectors, displacement vectors and position vectors In 2 or 3 dimensions.3.7.4Straight lines• understand the significance of all the symbols used when the equation of a straight line is expressed in the form r = a + tb, and find the equation of a line, given sufficient information e.g. finding the equation of a line given the position vector of a point on the line and a direction vector, or the position vectors of two points on the line.3.7.5Whether two lines are parallel• determine whether two lines are parallel, intersect or are skew, and find the point of intersection of two lines when it exists Calculation of the shortest distance between two skew lines is not required. Finding the equation of the common perpendicular to two skew lines is also not required.3.7.6Scalar product• use formulae to calculate the scalar product of two vectors, and use scalar products in problems involving lines and points. e.g. finding the angle between two lines, and finding the foot of the perpendicular from a point to a line; questions may involve 3D objects such as cuboids, tetrahedra (pyramids), etc. Knowledge of the vector product is not required.

Recognise the standard notations for one vector

Meaning 2D 3D
column components (xy)\begin{pmatrix}x\\y\end{pmatrix} (xyz)\begin{pmatrix}x\\y\\z\end{pmatrix}
basis-vector form xi+yjx\mathbf i+y\mathbf j xi+yj+zkx\mathbf i+y\mathbf j+z\mathbf k

AB→\overrightarrow{AB} is the directed vector from AA to BB; a bold/lowercase symbol such as a\mathbf a can name the same free vector. Equality means equal corresponding components.

A point P(x,y,z)P(x,y,z) names a location. Its position vector OP→\overrightarrow{OP} has the same numerical entries only because the origin and direction from OO are specified.

$\begin{pmatrix}2\\-1\\4\end{pmatrix}=2\mathbf i-\mathbf j+4\mathbf k$.

Use arrow/bold/column notation consistently. Magnitude and unit-vector calculations belong to objective 3, not to the notation definition.

Interpret component operations as geometric vector equations

Add/subtractcomponentwiseandscaleeverycomponent:Add/subtract componentwise and scale every component:(a_1,a_2,a_3)\pm(b_1,b_2,b_3),\qquad k\mathbf a.A negative $k$ reverses direction.

Directed displacements chain head-to-tail: AB→+BC→=AC→\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}. Opposite direction gives BA→=−AB→\overrightarrow{BA}=-\overrightarrow{AB}.

Geometry Position-vector equation
OABCOABC parallelogram OB→=OA→+OC→\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC} (with the stated vertex order)
midpoint MM of ABAB OM→=12(OA→+OB→)\overrightarrow{OM}=\tfrac12(\overrightarrow{OA}+\overrightarrow{OB})

If OA→=(2,1)\overrightarrow{OA}=(2,1) and OB→=(8,5)\overrightarrow{OB}=(8,5), midpoint MM has position (5,3)(5,3). Check by equal displacements AM→=MB→=(3,2)\overrightarrow{AM}=\overrightarrow{MB}=(3,2).

The general ratio theorem and scalar product are not included in this objective. Use only component operations, geometric chaining, parallelogram and midpoint.

Move between positions, displacements, magnitudes and unit directions

Quantity Formula
displacement AA to BB AB→=OB→−OA→\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}
magnitude of v=(v1,v2,v3)\mathbf v=(v_1,v_2,v_3) ∣v∣=v12+v22+v32|\mathbf v|=\sqrt{v_1^2+v_2^2+v_3^2}
unit vector along non-zero v\mathbf v v^=v/∣v∣\hat{\mathbf v}=\mathbf v/|\mathbf v|

Keep one origin for position vectors, subtract final minus initial for displacement, calculate its magnitude, then divide every component by that magnitude if a unit direction is required. The same workflow works in 2D or 3D.

From $A(1,-2,3)$ to $B(3,1,9)$:\overrightarrow{AB}=(2,3,6),\quad |\overrightarrow{AB}|=7,\quad \hat{\mathbf{AB}}=\tfrac17(2,3,6).

A unit vector must have magnitude 11. Reversing direction negates the unit vector but keeps the same magnitude.

AB→\overrightarrow{AB} is final minus initial; dividing by ∣v∣2|\mathbf v|^2 does not normalise the vector.

Construct a vector line from a point and a non-zero direction

Symbol in r=a+tb\mathbf r=\mathbf a+t\mathbf b Meaning
r\mathbf r position vector of a general point on the line
a\mathbf a position vector of one fixed point on the line
t∈Rt\in\mathbb R scalar parameter
b≠0\mathbf b\ne\mathbf0 direction vector

Given point AA and direction b\mathbf b, use r=OA→+tb\mathbf r=\overrightarrow{OA}+t\mathbf b. Given points A,BA,B, use direction AB→=OB→−OA→\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}.

Through $A(1,2,-1)$ and $B(4,0,5)$:\mathbf r=\begin{pmatrix}1\2\-1\end{pmatrix}+t\begin{pmatrix}3\-2\6\end{pmatrix}.

A different point on the same line or any non-zero scalar multiple of the direction produces an equivalent equation.

A point vector fixes location; a direction vector controls movement. Do not use A+BA+B as the direction between two points.

Classify two 3D lines by direction and component consistency

Check Conclusion
directions proportional parallel; test a point to distinguish coincident/distinct
directions not proportional and one parameter pair satisfies every component intersecting
directions not proportional and component equations are inconsistent skew

Use different parameters for the two lines. Equate position components, solve two equations, then verify the same parameter pair in the remaining component. If consistent, substitute to find the intersection point.

For $\mathbf r=\mathbf a+t\mathbf b$ and $\mathbf r=\mathbf c+s\mathbf d$, solve\mathbf a+t\mathbf b=\mathbf c+s\mathbf d.Avalid3Dintersectionrequiresallthreescalarequationssimultaneously.A valid 3D intersection requires all three scalar equations simultaneously.

If d=kb\mathbf d=k\mathbf b, test whether c−a\mathbf c-\mathbf a is also parallel to b\mathbf b. If yes the equations describe the same line; otherwise they are distinct parallel lines.

Two non-parallel 3D lines need not intersect. Shortest distance between skew lines and the equation of their common perpendicular are explicitly not required.

Use scalar products for line angles and perpendicular feet

\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta.Non-zero vectors are perpendicular exactly when their dot product is $0$.

For line directions b,d\mathbf b,\mathbf d, compute cos⁡θ=(b⋅d)/(∣b∣∣d∣)\cos\theta=(\mathbf b\cdot\mathbf d)/(|\mathbf b||\mathbf d|). If the smaller angle between lines is requested, use the acute value, equivalently ∣b⋅d∣|\mathbf b\cdot\mathbf d| in the numerator.

For line $\mathbf r=\mathbf a+t\mathbf b$ and point position $\mathbf p$, the foot is $\mathbf q=\mathbf a+t\mathbf b$ where(\mathbf p-\mathbf q)\cdot\mathbf b=0.Solve for $t$, then state $\mathbf q$.

This foot method works in 2D or 3D and in cuboid/tetrahedron contexts: the joining vector from the foot to the point must be perpendicular to the line direction.

The dot product is a scalar. Knowledge of the vector (cross) product is explicitly not required.