10 Vectors — 9MA0 spec coverage — Oscar Song Tutoring

10.1

Use vectors in two dimensions and in three dimensions.

A vector records magnitude and direction; in two or three dimensions it can be written in component form or with the unit vectors $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ .
Work component by component, keeping the $x$ , $y$ and $z$ entries aligned; subtract the initial point from the final point to form a displacement vector.
For example, from $A(1,2,3)$ to $B(4,0,5)$ the displacement is $\overrightarrow{AB}=(3,-2,2)$ .
A vector has no fixed location, whereas a point does; a common error is to confuse the coordinates of an endpoint with the components of the displacement leading to it.

Tier 1 · Easy

2 marks

ORIGINAL

The point $A$ has coordinates $(-2,5)$ and the point $B$ has coordinates $(4,1)$ . Find $\overrightarrow{AB}$ .

Tier 2 · Standard

2 marks

ORIGINAL

Given $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(-1,4,2)$ , find $2\mathbf{a}-\mathbf{b}$ .

Tier 3 · Hard

4 marks

ORIGINAL

Let $\mathbf{p}=(1,2,-1)$ , $\mathbf{q}=(2,-1,3)$ and $\mathbf{r}=(5,0,5)$ . Find scalars $\lambda$ and $\mu$ such that $\mathbf{r}=\lambda\mathbf{p}+\mu\mathbf{q}$ , and verify all three components.

10.2

Calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form.

For $\mathbf{v}=(a,b)$ , its magnitude is $|\mathbf{v}|=\sqrt{a^2+b^2}$ and a direction angle must be stated relative to a specified axis or bearing convention.
A vector of magnitude $r$ at angle $\theta$ anticlockwise from the positive $x$ -axis has components $(r\cos\theta,r\sin\theta)$ ; use signs or a quadrant-aware angle calculation when reversing the process.
For example, magnitude $10$ at $30^\circ$ above the positive $x$ -axis gives $(10\cos30^\circ,10\sin30^\circ)=(5\sqrt3,5)$ .
The value from $\tan^{-1}(b/a)$ alone can select the wrong quadrant; a common error is to report an acute reference angle without checking the component signs.

Tier 1 · Easy

3 marks

ORIGINAL

Find the magnitude and direction of the vector $(3,4)$ , giving the direction anticlockwise from the positive $x$ -axis to $1$ decimal place.

Tier 2 · Standard

3 marks

ORIGINAL

A vector has magnitude $14$ and direction $120^\circ$ anticlockwise from the positive $x$ -axis. Write it in exact component form.

Tier 3 · Hard

4 marks

ORIGINAL

A two-dimensional vector has magnitude $13$ , and the cosine of the angle it makes with the positive $x$ -axis is $\frac5{13}$ . Find all possible component forms and explain the ambiguity.

10.3

Add vectors diagrammatically and perform the algebraic operations of vector addition and multiplication by scalars, and understand their geometrical interpretations.

Vector addition combines successive displacements: placing the tail of $\mathbf{b}$ at the head of $\mathbf{a}$ makes the resultant from the first tail to the final head equal to $\mathbf{a}+\mathbf{b}$ .
Add corresponding components and multiply every component by a scalar; subtraction is addition of the opposite vector.
The vector $k\mathbf{a}$ is parallel to $\mathbf{a}$ , has magnitude $|k||\mathbf{a}|$ , and points in the reverse direction when $k<0$ .
A common error is to multiply only one component by a scalar or to draw vectors head-to-head instead of using a head-to-tail or parallelogram construction.

Tier 1 · Easy

1 mark

ORIGINAL

Given $\mathbf{u}=(2,-3)$ and $\mathbf{v}=(-5,4)$ , find $\mathbf{u}+\mathbf{v}$ .

Tier 2 · Standard

3 marks

ORIGINAL

Let $\mathbf{u}=(4,1)$ and $\mathbf{v}=(-2,5)$ . Find $2\mathbf{u}+\mathbf{v}$ and describe a head-to-tail construction for this resultant.

Tier 3 · Hard

4 marks

ORIGINAL

The vectors $\mathbf{a}=(1,2)$ and $\mathbf{b}=(3,-1)$ combine to give $\mathbf{w}=(11,1)$ . Find $\alpha$ and $\beta$ such that $\mathbf{w}=\alpha\mathbf{a}+\beta\mathbf{b}$ , and interpret the result geometrically.

10.4

Understand and use position vectors; calculate the distance between two points represented by position vectors.

The position vector of a point $A$ is $\overrightarrow{OA}$ from a fixed origin $O$ ; if these vectors are $\mathbf{a}$ and $\mathbf{b}$ , then $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .
Find a displacement by subtracting position vectors in final-minus-initial order, then find distance by taking the magnitude of that displacement.
A point dividing $AB$ internally in the fraction $t$ from $A$ to $B$ has position vector $\mathbf{a}+t(\mathbf{b}-\mathbf{a})$ .
Distance is a non-negative scalar, not a vector; a common error is to quote $\mathbf{b}-\mathbf{a}$ as the distance without calculating its magnitude.

Tier 1 · Easy

3 marks

ORIGINAL

Points $A$ and $B$ have position vectors $(2,-1)$ and $(7,3)$ . Find $\overrightarrow{AB}$ and the exact distance $AB$ .

Tier 2 · Standard

3 marks

ORIGINAL

The position vectors of $A$ and $B$ are $(-1,2,4)$ and $(3,-4,7)$ . Calculate the exact distance $AB$ .

Tier 3 · Hard

5 marks

ORIGINAL

Points $A$ and $B$ have position vectors $(1,-2,3)$ and $(9,6,-1)$ . The point $M$ divides $AB$ internally in the ratio $AM:MB=3:1$ . Find the position vector of $M$ and the exact distance $OM$ .

10.5

Use vectors to solve problems in pure mathematics and in context (including forces).

Vector models turn geometrical displacements or forces into component equations; equilibrium means that the vector sum of all forces is zero.
Choose and state positive coordinate directions, resolve every vector consistently, then equate components or use position-vector relationships.
In a parallelogram with adjacent position vectors $\mathbf{a}$ and $\mathbf{b}$ , the opposite vertex has position vector $\mathbf{a}+\mathbf{b}$ and both diagonals share midpoint $\frac12(\mathbf{a}+\mathbf{b})$ .
A common error is to balance force magnitudes without balancing directions; equal numerical magnitudes do not guarantee equilibrium unless the vector sum is zero.

Tier 1 · Easy

3 marks

ORIGINAL

Two forces acting on a particle are $\mathbf{F}_1=(4,-1)$ N and $\mathbf{F}_2=(-2,5)$ N. Find the resultant force and its magnitude.

Tier 2 · Standard

3 marks

ORIGINAL

The points $O,A,B,C$ have position vectors $\mathbf{0},\mathbf{a},\mathbf{b},\mathbf{a}+\mathbf{b}$ respectively. Use vectors to prove that the diagonals $OC$ and $AB$ bisect each other.

Tier 3 · Hard

5 marks

ORIGINAL

Three vectors sum to the zero vector. Two of them are $\mathbf{a}=(3,4,-2)$ and $\mathbf{b}=(-5,1,6)$ . Find the third vector $\mathbf{c}$ , its exact magnitude and a unit vector in its direction.