10 Vectors — printable coverage pack

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

1. The point $A$ has coordinates $(-2,5)$ and the point $B$ has coordinates $(4,1)$ . Find $\overrightarrow{AB}$ .[2 marks]
Answer
- $\overrightarrow{AB}=(6,-4)$
Method: Subtract the coordinates of $A$ from those of $B$ : $\overrightarrow{AB}=(4-(-2),1-5)=(6,-4)$ .

Tier 2 · Standard

1. Given $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(-1,4,2)$ , find $2\mathbf{a}-\mathbf{b}$ .[2 marks]
Answer
- $2\mathbf{a}-\mathbf{b}=(5,-6,4)$
Method: First $2\mathbf{a}=(4,-2,6)$ . Subtract corresponding components of $\mathbf{b}$ to get $(4-(-1),-2-4,6-2)=(5,-6,4)$ .

Tier 3 · Hard

1. 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.[4 marks]
Answer
- $\lambda=1$ and $\mu=2$
Method: Equating components gives $\lambda+2\mu=5$ , $2\lambda-\mu=0$ and $-\lambda+3\mu=5$ . The second equation gives $\mu=2\lambda$ ; substituting in the first gives $5\lambda=5$ , so $\lambda=1$ and $\mu=2$ . The third component checks: $-1+3(2)=5$ .

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

1. Find the magnitude and direction of the vector $(3,4)$ , giving the direction anticlockwise from the positive $x$ -axis to $1$ decimal place.[3 marks]
Answer
- Magnitude $5$ ; direction $53.1^\circ$
Method: The magnitude is $\sqrt{3^2+4^2}=5$ . Both components are positive, so the vector is in the first quadrant and $\theta=\tan^{-1}(4/3)=53.1^\circ$ .

Tier 2 · Standard

1. A vector has magnitude $14$ and direction $120^\circ$ anticlockwise from the positive $x$ -axis. Write it in exact component form.[3 marks]
Answer
- $(-7,7\sqrt3)$
Method: Use $(14\cos120^\circ,14\sin120^\circ)$ . Since $\cos120^\circ=-\frac12$ and $\sin120^\circ=\frac{\sqrt3}{2}$ , the vector is $(-7,7\sqrt3)$ .

Tier 3 · Hard

1. 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.[4 marks]
Answer
- $(5,12)$ or $(5,-12)$
Method: The horizontal component is $13(5/13)=5$ . If the vertical component is $y$ , then $5^2+y^2=13^2$ , so $y^2=144$ and $y=\pm12$ . Cosine fixes the positive horizontal component but does not distinguish an angle above the axis from its reflection below the axis.

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. Given $\mathbf{u}=(2,-3)$ and $\mathbf{v}=(-5,4)$ , find $\mathbf{u}+\mathbf{v}$ .[1 mark]
Answer
- $(-3,1)$
Method: Add corresponding components: $\mathbf{u}+\mathbf{v}=(2+(-5),-3+4)=(-3,1)$ .

Tier 2 · Standard

1. 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.[3 marks]
Answer
- $2\mathbf{u}+\mathbf{v}=(6,7)$ .
- Place two copies of $\mathbf{u}$ and then one copy of $\mathbf{v}$ head-to-tail; the resultant joins the initial tail to the final head.
Method: Calculate $2\mathbf{u}=(8,2)$ , then add $\mathbf{v}$ to obtain $(8-2,2+5)=(6,7)$ . Diagrammatically, translate the vectors without rotating them and place the second $\mathbf{u}$ after the first, followed by $\mathbf{v}$ ; the direct closing vector is the sum.

Tier 3 · Hard

1. 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.[4 marks]
Answer
- $\alpha=2$ and $\beta=3$ , so $\mathbf{w}=2\mathbf{a}+3\mathbf{b}$ .
Method: Equating components gives $\alpha+3\beta=11$ and $2\alpha-\beta=1$ . From the second, $\beta=2\alpha-1$ ; substitution gives $7\alpha=14$ , hence $\alpha=2$ and $\beta=3$ . Geometrically, two copies of $\mathbf{a}$ and three copies of $\mathbf{b}$ placed head-to-tail have resultant $\mathbf{w}$ .

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

1. Points $A$ and $B$ have position vectors $(2,-1)$ and $(7,3)$ . Find $\overrightarrow{AB}$ and the exact distance $AB$ .[3 marks]
Answer
- $\overrightarrow{AB}=(5,4)$ and $AB=\sqrt{41}$
Method: Subtract the position vector of $A$ from that of $B$ : $\overrightarrow{AB}=(7-2,3-(-1))=(5,4)$ . Therefore $AB=\sqrt{5^2+4^2}=\sqrt{41}$ .

Tier 2 · Standard

1. The position vectors of $A$ and $B$ are $(-1,2,4)$ and $(3,-4,7)$ . Calculate the exact distance $AB$ .[3 marks]
Answer
- $AB=\sqrt{61}$
Method: $\overrightarrow{AB}=(3-(-1),-4-2,7-4)=(4,-6,3)$ . Hence $AB=|\overrightarrow{AB}|=\sqrt{4^2+(-6)^2+3^2}=\sqrt{61}$ .

Tier 3 · Hard

1. 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 $OM$ .[5 marks]
Answer
- $\overrightarrow{OM}=(7,4,0)$ and $OM=\sqrt{65}$
Method: Since $AM$ is three of the four equal ratio parts, $\overrightarrow{OM}=\overrightarrow{OA}+\frac34(\overrightarrow{OB}-\overrightarrow{OA})$ . Now $\overrightarrow{OB}-\overrightarrow{OA}=(8,8,-4)$ , so $\overrightarrow{OM}=(1,-2,3)+(6,6,-3)=(7,4,0)$ . Thus $OM=\sqrt{7^2+4^2}=\sqrt{65}$ .

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

1. 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.[3 marks]
Answer
- Resultant $(2,4)$ N; magnitude $2\sqrt5$ N
Method: Add the force components: $\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2=(4-2,-1+5)=(2,4)$ N. Its magnitude is $|\mathbf{R}|=\sqrt{2^2+4^2}=\sqrt{20}=2\sqrt5$ N.

Tier 2 · Standard

1. 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.[3 marks]
Answer
- Both diagonals have midpoint position vector $\frac12(\mathbf{a}+\mathbf{b})$ , so they bisect each other.
Method: The midpoint of $OC$ has position vector $\frac12[\mathbf{0}+(\mathbf{a}+\mathbf{b})]=\frac12(\mathbf{a}+\mathbf{b})$ . The midpoint of $AB$ has position vector $\frac12(\mathbf{a}+\mathbf{b})$ as well. Since the two diagonals have the same midpoint, each bisects the other.

Tier 3 · Hard

1. 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.[5 marks]
Answer
- $\mathbf{c}=(2,-5,-4)$
- $|\mathbf{c}|=3\sqrt5$ and a unit direction vector is $\frac{1}{3\sqrt5}(2,-5,-4)$ .
Method: The condition $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ gives $\mathbf{c}=-(\mathbf{a}+\mathbf{b})$ . The first two vectors sum to $(-2,5,4)$ , so $\mathbf{c}=(2,-5,-4)$ . Its magnitude is $\sqrt{2^2+(-5)^2+(-4)^2}=\sqrt{45}=3\sqrt5$ . Dividing the vector by this magnitude gives the unit vector $\frac{1}{3\sqrt5}(2,-5,-4)$ .