This Year 13 vectors in three dimensions revision notes is written for pupils aged 17-18 and follows the National Curriculum. In short, it shows you how to use three-dimensional vectors, magnitude and geometric problems. Before you start, it helps to revise integration techniques and differential equations and numerical methods, because this lesson builds on them.
Year 13 vectors in three dimensions revision notes: what you will learn
In two dimensions a vector has two components. In three dimensions we add a third axis, the \(z\)-axis, and a third unit vector \(\mathbf k\).
Starter activity
Stand in a corner of your classroom and imagine a straight line to the opposite corner of the ceiling. How could you calculate its length from the dimensions of the room? Draw a sketch showing a right-angled triangle on the floor and a second right-angled triangle that stands upright. Explain how Pythagoras is used twice.
Vectors in three dimensions
Definition: 3D vector and position vector
Every vector can be written \(\mathbf a=x\mathbf i+y\mathbf j+z\mathbf k\) or as a column \(\begin{pmatrix}x\\y\\z\end{pmatrix}\). The unit vectors \(\mathbf i\), \(\mathbf j\), \(\mathbf k\) have length 1 and point along the three perpendicular axes. The position vector of \(P(x,y,z)\) is \(\overrightarrow{OP}=x\mathbf i+y\mathbf j+z\mathbf k\).
Rule: adding and scaling
Add and subtract component by component; a scalar multiplies every component.
Worked example 1
\(\mathbf a=2\mathbf i+3\mathbf j-\mathbf k\) and \(\mathbf b=\mathbf i-\mathbf j+4\mathbf k\). Find \(\mathbf a+\mathbf b\) and \(2\mathbf a-\mathbf b\).
\(\mathbf a+\mathbf b=3\mathbf i+2\mathbf j+3\mathbf k\) and \(2\mathbf a-\mathbf b=3\mathbf i+7\mathbf j-6\mathbf k\).
Magnitude and unit vectors
Rule: magnitude and unit vector
By applying Pythagoras twice, \(|x\mathbf i+y\mathbf j+z\mathbf k|=\sqrt{x^2+y^2+z^2}\). A unit vector in the direction of \(\mathbf a\) is \(\hat{\mathbf a}=\dfrac{\mathbf a}{|\mathbf a|}\). For a vector of length \(L\) in that direction, multiply the unit vector by \(L\).
Worked example 2
Find \(|3\mathbf i-4\mathbf j+12\mathbf k|\), the unit vector, and a vector of magnitude 26 in this direction.
\(\sqrt{9+16+144}=13\). The unit vector is \(\tfrac1{13}(3\mathbf i-4\mathbf j+12\mathbf k)\), and the vector of magnitude 26 is \(6\mathbf i-8\mathbf j+24\mathbf k\).
Worked example 3
If \(|2\mathbf i+\lambda\mathbf j+3\mathbf k|=7\), find \(\lambda\).
\(4+\lambda^2+9=49\), so \(\lambda^2=36\) and \(\lambda=\pm6\).
Vectors between points
Rule: vector, distance and midpoint
\(\overrightarrow{AB}=\mathbf b-\mathbf a\): always "end minus start". The distance is \(|\overrightarrow{AB}|\), the midpoint has position vector \(\tfrac12(\mathbf a+\mathbf b)\), and the point dividing \(AB\) in ratio \(m:n\) is \(\mathbf a+\dfrac m{m+n}(\mathbf b-\mathbf a)\).
Worked example 4
\(A(1,2,3)\), \(B(4,-2,15)\). Find \(AB\), the midpoint and the point \(P\) with \(AP:PB=1:2\).
\(|\overrightarrow{AB}|=|3\mathbf i-4\mathbf j+12\mathbf k|=13\). Midpoint \(\left(\tfrac52,0,9\right)\). \(\mathbf p=\mathbf a+\tfrac13(\mathbf b-\mathbf a)=(2,\tfrac23,7)\).
Parallel vectors and collinear points
Definition: parallel and collinear
Vectors are parallel if \(\mathbf u=k\mathbf v\). Points \(A,B,C\) are collinear if \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are parallel: they share \(B\), so all three lie on one line.
Worked example 5
Show that \(A(1,2,3)\), \(B(3,5,7)\), \(C(7,11,15)\) are collinear.
\(\overrightarrow{AB}=2\mathbf i+3\mathbf j+4\mathbf k\) and \(\overrightarrow{BC}=4\mathbf i+6\mathbf j+8\mathbf k=2\overrightarrow{AB}\). They are parallel and share \(B\), so the points are collinear with \(AB:BC=1:2\).
Worked example 6
Find \(p\) and \(q\) so that \(p\mathbf i+6\mathbf j+9\mathbf k\) is parallel to \(2\mathbf i+q\mathbf j+3\mathbf k\).
Equate ratios: \(\dfrac p2=\dfrac6q=\dfrac93=3\), so \(p=6\) and \(q=2\).
Geometric problems
Method: geometry with 3D vectors
- Parallelogram \(ABCD\): use \(\overrightarrow{AD}=\overrightarrow{BC}\), so \(\mathbf d=\mathbf a+\mathbf c-\mathbf b\).
- Classify a triangle by calculating the three side lengths, and test a right angle with Pythagoras.
- Use the cosine rule for an angle: \(\cos B=\dfrac{AB^2+BC^2-AC^2}{2\,AB\cdot BC}\).
Worked example 7
\(A(1,0,2)\), \(B(3,1,4)\), \(C(3,2,1)\). Classify the triangle and find \(\angle ABC\).
\(AB=3\), \(AC=3\), \(BC=\sqrt{10}\): isosceles. \(\cos B=\dfrac{9+10-9}{2\times3\times\sqrt{10}}=0.527\), so \(\angle ABC=58.2^\circ\).
Worked example 8
Show that \(O(0,0,0)\), \(B(2,2,1)\), \(C(-2,1,2)\) form a right angle at \(O\). Find the space diagonal of a 3 by 4 by 12 cuboid.
\(OB^2+OC^2=9+9=18=BC^2\), so \(\angle O=90^\circ\). The space diagonal is \(\sqrt{9+16+144}=13\), at \(\tan^{-1}\tfrac{12}{5}=67.4^\circ\) to the base.
Vectors in mechanics
Worked example 9
Forces \(\mathbf F_1=(2\mathbf i+3\mathbf j-\mathbf k)\) N and \(\mathbf F_2=(\mathbf i-\mathbf j+4\mathbf k)\) N act on a particle in equilibrium with \(\mathbf F_3\). Find \(\mathbf F_3\) and its magnitude.
\(\mathbf F_1+\mathbf F_2+\mathbf F_3=\mathbf 0\) gives \(\mathbf F_3=(-3\mathbf i-2\mathbf j-3\mathbf k)\) N, of magnitude \(\sqrt{22}=4.69\) N.
Worked example 10
A particle has position \(\mathbf r=(1+2t)\mathbf i+(2-t)\mathbf j+(3+2t)\mathbf k\). Find its velocity and speed.
The velocity is the constant \(2\mathbf i-\mathbf j+2\mathbf k\) and the speed is 3 m/s. Two particles collide if they are at the same position at the same time: equate all three components for the same \(t\).
Watch out
\(\overrightarrow{AB}=\mathbf b-\mathbf a\), not \(\mathbf a-\mathbf b\). Do not forget the \(z\)-component in a magnitude. Give the vector between points, not a position vector, when asked. Check all three components for collinearity or collision.
Worked example 11: Direction vectors of a given length
A drone is at \(A(1,2,3)\) and flies towards \(B(7,5,5)\) at a speed of 14 m/s. Write its velocity as a vector.
\(\overrightarrow{AB}=6\mathbf i+3\mathbf j+2\mathbf k\) with magnitude \(\sqrt{36+9+4}=7\). The unit vector is \(\tfrac17(6\mathbf i+3\mathbf j+2\mathbf k)\). Multiply by the speed: \(\mathbf v=14\times\tfrac17(6,3,2)=12\mathbf i+6\mathbf j+4\mathbf k\) m/s. Check: \(\sqrt{144+36+16}=14\).
Worked example 12: Finding a point on a line
Find the coordinates of the point on the line from \(A(2,0,1)\) to \(B(8,6,4)\) where the third coordinate is 3.
\(\overrightarrow{AB}=(6,6,3)\). A general point is \(\mathbf a+\lambda\overrightarrow{AB}=(2+6\lambda,6\lambda,1+3\lambda)\). The third coordinate is 3, so \(1+3\lambda=3\) and \(\lambda=\tfrac23\). The point is \((2+4,4,3)=(6,4,3)\). Since \(0\lt \lambda<1\) the point lies between \(A\) and \(B\).
Worked example 13: Proving a right angle with Pythagoras
Show that triangle \(PQR\) with \(P(1,1,1)\), \(Q(3,2,3)\) and \(R(2,3,1)\) is right-angled.
\(PQ^2=4+1+4=9\), \(PR^2=1+4+0=5\) and \(QR^2=1+1+4=6\). Check which side is largest: \(PQ^2=9\). Is \(PR^2+QR^2=5+6=11\)? No. The sides are not a Pythagorean triple, so the triangle is not right-angled. Always test the largest side squared against the sum of the other two squares.
Worked example 14: A moving particle reaching a plane
A particle moves with \(\mathbf r=(2+t)\mathbf i+(1-t)\mathbf j+(12-3t)\mathbf k\). Find when and where it reaches the ground, \(z=0\).
\(12-3t=0\) gives \(t=4\). Substituting, \(\mathbf r=6\mathbf i-3\mathbf j+0\mathbf k\), so the particle lands at \((6,-3,0)\). Its speed is \(\sqrt{1+1+9}=\sqrt{11}=3.32\) m/s and the distance travelled is \(4\sqrt{11}=13.3\) m.
Worked example 15: Collision of two particles with a parameter
Particles \(P\) and \(Q\) have position vectors \(\mathbf r_P=(2+t)\mathbf i+(3-t)\mathbf j+5\mathbf k\) and \(\mathbf r_Q=(8-2t)\mathbf i+(p+t)\mathbf j+(5)\mathbf k\). They collide. Find \(p\) and the time of collision.
The \(\mathbf i\) components give \(2+t=8-2t\), so \(t=2\). The \(\mathbf j\) components must also be equal at \(t=2\): \(3-2=p+2\), so \(p=-1\). The \(\mathbf k\) components are both 5. The collision occurs at \(t=2\) at the point \((4,1,5)\).
Note: the axes and the right-hand rule
The three axes \(x\), \(y\), \(z\) are perpendicular and form a right-handed set: if you curl the fingers of your right hand from the \(x\)-axis towards the \(y\)-axis, your thumb points along the \(z\)-axis. In most questions \(\mathbf k\) points vertically upwards, so the \(z\)-coordinate is the height above the origin. Always read the question carefully to see which direction is "up". Distances in a vertical plane combine with horizontal distances using Pythagoras, so the space diagonal of a cuboid with edges \(a\), \(b\), \(c\) has length \(\sqrt{a^2+b^2+c^2}\).
Worked example 16: Finding an unknown coordinate from a distance
The point \(P\) has coordinates \((3,-1,p)\) and the origin is \(O\). Given that \(OP=\sqrt{35}\), find the possible values of \(p\).
\(|\overrightarrow{OP}|^2=9+1+p^2=35\), so \(p^2=25\) and \(p=\pm5\). There are two possible points, \((3,-1,5)\) and \((3,-1,-5)\), one above and one below the \(xy\)-plane. When a question says "find the possible values", remember the negative root.
Worked example 17: Using a ratio to find an endpoint
\(M(3,4,5)\) is the midpoint of \(AB\) and \(A\) is \((1,0,2)\). Find \(B\).
The midpoint formula gives \(\mathbf m=\tfrac12(\mathbf a+\mathbf b)\), so \(\mathbf b=2\mathbf m-\mathbf a=(6,8,10)-(1,0,2)=(5,8,8)\). Check: the midpoint of \((1,0,2)\) and \((5,8,8)\) is \((3,4,5)\). Equivalently, \(\overrightarrow{AM}=(2,4,3)\) and \(\overrightarrow{MB}\) must be the same vector, so \(B=M+(2,4,3)=(5,8,8)\).
Worked example 18: Finding the angle between a line and a plane
A ski lift cable runs from \(A(0,0,10)\) to \(B(60,80,60)\) (metres, \(\mathbf k\) upwards). Find the length of the cable and its angle of elevation.
\(\overrightarrow{AB}=(60,80,50)\). Length \(=\sqrt{3600+6400+2500}=\sqrt{12500}=111.8\) m. The horizontal distance is \(\sqrt{60^2+80^2}=100\) m and the rise is 50 m, so \(\tan\theta=\dfrac{50}{100}=0.5\) and \(\theta=26.6^\circ\). The cable rises by 1 m for every 2 m horizontally.
Worked example 19: Showing that a triangle is isosceles
Show that triangle \(ABC\) with \(A(1,1,1)\), \(B(3,3,2)\) and \(C(3,-1,2)\) is isosceles.
\(AB=\sqrt{4+4+1}=3\). \(AC=\sqrt{4+4+1}=3\). \(BC=\sqrt{0+16+0}=4\). Two sides are equal, so the triangle is isosceles with apex \(A\). Angle at \(A\): \(\cos A=\dfrac{9+9-16}{2\times3\times3}=\dfrac19\), so \(A=83.6^\circ\).
Watch out: common errors with vectors
- Subtracting in the wrong order: \(\overrightarrow{AB}=\mathbf b-\mathbf a\), not \(\mathbf a-\mathbf b\). A sketch showing the direction of the arrow prevents this.
- Forgetting the third component when finding a magnitude, or forgetting the square root.
- Claiming that points are collinear because two vectors are parallel without noting that they share a common point.
- Equating position vectors at different times: for a collision the same value of \(t\) must satisfy every component.
- Leaving a vector answer as a pair of numbers when a point or a scalar is requested. Read whether the question asks for a position vector, a vector, a coordinate pair or a distance.
Worked example 20: Showing that two lines of motion cross but particles do not collide
Particle \(A\) has position \(\mathbf r_A=(t)\mathbf i+(2t)\mathbf j+(3)\mathbf k\) and particle \(B\) has position \(\mathbf r_B=(4-t)\mathbf i+(2)\mathbf j+(3)\mathbf k\). Do they collide?
Equate the \(\mathbf i\) components: \(t=4-t\), so \(t=2\). Equate the \(\mathbf j\) components: \(2t=2\), so \(t=1\). The two values differ, so the particles are never at the same point at the same time. However, the paths do cross at \((1,2,3)\): particle \(A\) is there at \(t=1\), and \(B\) is there at \(t=3\). The paths meet but the particles arrive at different times, so there is no collision.
Method: solving geometry problems in three dimensions
- Write the position vector of every point and label the vectors you need as "end minus start".
- Use the magnitude formula \(\sqrt{x^2+y^2+z^2}\) for lengths, and the cosine rule for angles in triangles.
- For parallelism or collinearity, write one vector as a scalar multiple of another.
- For motion, equate all three components at the same time for a collision; minimise \(d^2\), not \(d\), to find a closest approach.
- Finish by answering in the context, with units.
Note: why minimise the square of the distance
The distance \(d=\sqrt{(\ldots)}\) is awkward to differentiate, but \(d^2\) is a quadratic in \(t\) and \(d\) is smallest exactly when \(d^2\) is smallest, because the square root is an increasing function. So find \(t\) from \(\dfrac{d(d^2)}{dt}=0\) and then take the square root at the end.
Did you know?
The three unit vectors \(\mathbf i\), \(\mathbf j\), \(\mathbf k\) were introduced in the 1840s by the Irish mathematician William Rowan Hamilton, who carved the key formula for his quaternions into Broom Bridge in Dublin as he walked past. Today three-dimensional vectors are used by computer-game designers to move characters and cameras, and by air-traffic controllers to keep aircraft apart.
Otto's tip
Draw a quick sketch and label the points before you calculate. It stops sign errors with end-minus-start.
How to use this Year 13 vectors in three dimensions revision notes
First, read the key points so that you know where the Year 13 vectors in three dimensions revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the vectors in three dimensions worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the vectors in three dimensions test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 13 vectors in three dimensions revision notes gives you everything you need for the chapter.
To sum up, regular practice works better than one long session. For instance, ten minutes a day is often enough. Besides, mistakes are useful, because they show what to revise next. Then, once a method feels easy, try a harder question. As a result, you build confidence step by step.
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