This University vectors and systems of linear equations revision notes is written for pupils aged 18-19. In short, it shows you how to solve linear systems by row reduction. Before you start, it helps to revise integration and the fundamental theorem of calculus and sequences and series convergence, because this lesson builds on them.
University vectors and systems of linear equations revision notes: what you will learn
Vectors describe direction and magnitude, and systems of linear equations are solved by systematic row operations. Together they are the starting point of linear algebra.
Starter activity
A builder needs to know whether three steel beams, each described by a straight line in space, actually meet at one point, and a surveyor wants the height of a sloping roof above a given spot. Both questions come down to equations of lines and planes. Before reading on, write down the equation of the floor \(z=0\) and of a wall, and ask yourself: how many equations do you need to pin down a single point in space?
Vectors in R^n
Definition: dot product and length
A vector in \(\mathbb R^3\) is an ordered triple \(\mathbf a=(a_1,a_2,a_3)\), added and scaled componentwise. The length is \(|\mathbf a|=\sqrt{a_1^2+a_2^2+a_3^2}\). The dot product is \(\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta\).
Rule: dot product results
\(\mathbf a\perp\mathbf b\) exactly when \(\mathbf a\cdot\mathbf b=0\). The Cauchy-Schwarz inequality \(|\mathbf a\cdot\mathbf b|\le|\mathbf a||\mathbf b|\) holds for all vectors. The projection of \(\mathbf a\) onto \(\mathbf b\) is \(\frac{\mathbf a\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\mathbf b\).
Example 1
Find the angle between \(\mathbf a=(1,2,3)\) and \(\mathbf b=(2,-1,4)\).
\(\mathbf a\cdot\mathbf b=2-2+12=12\), \(|\mathbf a|=\sqrt{14}\), \(|\mathbf b|=\sqrt{21}\). So \(\cos\theta=\frac{12}{\sqrt{294}}\approx0.700\) and \(\theta\approx45.6^\circ\).
The cross product
Definition: cross product
In \(\mathbb R^3\), \(\mathbf a\times\mathbf b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1)\). It is perpendicular to both vectors, and \(|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta\) is the area of the parallelogram they span.
Property: cross product facts
\(\mathbf a\times\mathbf b=-\mathbf b\times\mathbf a\), and \(\mathbf a\times\mathbf b=\mathbf 0\) if and only if the vectors are parallel. The scalar triple product \(\mathbf a\cdot(\mathbf b\times\mathbf c)\) gives the signed volume of a parallelepiped.
Example 2
Find \(\mathbf a\times\mathbf b\) for \(\mathbf a=(1,2,3)\) and \(\mathbf b=(2,-1,4)\).
\(\mathbf a\times\mathbf b=(8+3,\ 6-4,\ -1-4)=(11,2,-5)\) and \(|\mathbf a\times\mathbf b|=\sqrt{150}=5\sqrt6\).
Lines and planes
Rule: equations of lines and planes
A line through \(\mathbf p\) with direction \(\mathbf d\) is \(\mathbf r=\mathbf p+t\mathbf d\). A plane with normal \(\mathbf n\) through \(\mathbf p\) is \(\mathbf n\cdot\mathbf r=\mathbf n\cdot\mathbf p\). The distance from the origin to \(\mathbf n\cdot\mathbf r=d\) is \(\frac{|d|}{|\mathbf n|}\). For the plane through three points, use the cross product of two edge vectors as the normal.
Example 3
Find the plane with normal \(\mathbf n=(2,-1,4)\) through \((1,2,3)\) and its distance from the origin.
\(2x-y+4z=2-2+12=12\). The distance is \(\frac{12}{\sqrt{21}}\approx2.62\).
Systems of linear equations
Definition: augmented matrix and row operations
A system of \(m\) linear equations in \(n\) unknowns can be written \(A\mathbf x=\mathbf b\), with augmented matrix \([A\mid\mathbf b]\). The elementary row operations are: swap two rows; multiply a row by a non-zero constant; add a multiple of one row to another. They do not change the solution set.
Method: Gaussian elimination
- Write the augmented matrix.
- Use row operations to make zeros below each leading entry (row echelon form).
- Back-substitute from the bottom row up, or continue to reduced row echelon form (leading 1s, zeros above and below) to read off the answer.
- Substitute the solution into every original equation.
Example 4
Solve \(x+y+z=6\), \(2x-y+z=3\), \(x+2y-z=2\).
\(R_2\to R_2-2R_1\) and \(R_3\to R_3-R_1\) give \(-3y-z=-9\) and \(y-2z=-4\). From the second, \(y=2z-4\); substituting, \(-6z+12-z=-9\), so \(z=3\), \(y=2\), \(x=1\). The unique solution is \((1,2,3)\).
How many solutions?
Rule: number of solutions
A system has either no solution, exactly one, or infinitely many. The rank of \(A\) is the number of non-zero rows in echelon form. The system is inconsistent if the echelon form contains a row \([0\ 0\ \cdots\ 0\mid c]\) with \(c\ne0\). If it is consistent and rank equals the number of unknowns, the solution is unique; if rank is less, there are \(n-\text{rank}\) free parameters.
Property: homogeneous systems
A homogeneous system \(A\mathbf x=\mathbf 0\) always has the trivial solution, and has non-trivial solutions if and only if rank \(\lt n\).
Example 5
Discuss \(x+y=1,\ 2x+2y=3\), and solve \(x+y+z=1,\ x-y+2z=0\).
The first is inconsistent: the second equation minus twice the first gives \(0=1\). The second has rank 2 with 3 unknowns, so it is a line: \(\left(\tfrac12-\tfrac32t,\ \tfrac12+\tfrac12t,\ t\right)\).
Watch out
- Using the dot product where a cross product is needed.
- Forgetting that the cross product is anticommutative.
- Using a step that is not an elementary row operation, such as adding a constant to a row.
- Failing to check for inconsistency before back-substitution.
Geometry and linear systems together
Key fact: distances in three dimensions
Three formulas cover most distance problems. The distance from a point \(P\) to the plane \(ax+by+cz=d\) is \(\dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}\). The distance from \(P\) to the line through \(A\) with direction \(\mathbf d\) is \(\dfrac{|\overrightarrow{AP}\times\mathbf d|}{|\mathbf d|}\), because the cross product gives the area of a parallelogram whose base is \(|\mathbf d|\). The distance between skew lines is \(\dfrac{|\overrightarrow{AB}\cdot(\mathbf d_1\times\mathbf d_2)|}{|\mathbf d_1\times\mathbf d_2|}\): the projection of any connecting vector onto the common perpendicular.
Method: solving and classifying a linear system
- Write the augmented matrix \((A\mid\mathbf b)\).
- Use row operations (swap rows, multiply a row by a non-zero number, add a multiple of one row to another) to reach row echelon form.
- If a row reads \((0\ \cdots\ 0\mid c)\) with \(c\ne0\), the system is inconsistent.
- Otherwise the number of free parameters is the number of unknowns minus the rank.
- Solve by back substitution, write the solution in vector form, and check it in every original equation.
Worked example 6: The angle between two planes
Find the acute angle between the planes \(2x-y+2z=3\) and \(x+2y+2z=7\).
The angle between two planes equals the angle between their normals \(\mathbf n_1=(2,-1,2)\) and \(\mathbf n_2=(1,2,2)\). We have \(\mathbf n_1\cdot\mathbf n_2=2-2+4=4\) and \(|\mathbf n_1|=|\mathbf n_2|=3\). So \(\cos\theta=\dfrac49\) and \(\theta\approx63.6^\circ\). If the dot product had been negative we would take its absolute value to obtain the acute angle.
Worked example 7: A line of intersection
Find a vector equation of the line where \(x+y-z=1\) and \(2x-y+z=5\) meet.
Adding the equations gives \(3x=6\), so \(x=2\) on the whole line. Then \(y-z=-1\), so taking \(z=t\) we get \(y=t-1\). The line is \(\mathbf r=(2,-1,0)+t(0,1,1)\). As a check, the direction should be perpendicular to both normals: \((0,1,1)\cdot(1,1,-1)=0\) and \((0,1,1)\cdot(2,-1,1)=0\). Equivalently, the direction is a multiple of \((1,1,-1)\times(2,-1,1)=(0,-3,-3)\).
Worked example 8: A system in context
A garden centre sells three types of compost bag with masses 10 kg, 20 kg and 40 kg. A delivery of 30 bags weighs 700 kg and contains twice as many 10 kg bags as 40 kg bags. How many of each are there?
Let \(x\), \(y\), \(z\) be the numbers of 10, 20 and 40 kg bags. Then \(x+y+z=30\), \(10x+20y+40z=700\) and \(x=2z\). Substituting the last equation: \(3z+y=30\) and \(20z+20y+40z=700\), i.e. \(60z+20y=700\), so \(3z+y=35\). These two equations contradict each other, so no such delivery exists. Changing the total mass to 600 kg gives \(3z+y=30\) twice: infinitely many solutions in theory, but only whole numbers with \(y=30-3z\ge0\) make sense, such as \(z=5\), \(x=10\), \(y=15\).
Worked example 9: Projection and the component perpendicular
Split \(\mathbf v=(3,4,5)\) into a part parallel to \(\mathbf u=(1,0,1)\) and a part perpendicular to it.
The parallel part is \(\dfrac{\mathbf v\cdot\mathbf u}{|\mathbf u|^2}\mathbf u=\dfrac82(1,0,1)=(4,0,4)\). The perpendicular part is \(\mathbf v-(4,0,4)=(-1,4,1)\), and indeed \((-1,4,1)\cdot(1,0,1)=0\). This decomposition is used in mechanics to split a force into components along and across a slope.
Worked example 10: Rank and the number of solutions
For which values of \(k\) does \(x+y+kz=1\), \(x+ky+z=1\), \(kx+y+z=1\) have a unique solution?
The determinant of the coefficient matrix is \(-(k-1)^2(k+2)\). For \(k\ne1\) and \(k\ne-2\) it is non-zero and the solution is unique: by symmetry \(x=y=z=\dfrac1{k+2}\). For \(k=1\) all three equations are \(x+y+z=1\): rank 1, two free parameters, a whole plane of solutions. For \(k=-2\), adding the three equations gives \(0=3\): no solution.
Worked example 11: Area of a triangle in space
Find the area of the triangle with vertices \(A(1,1,0)\), \(B(3,2,1)\) and \(C(2,4,2)\), and the angle at \(A\).
The sides from \(A\) are \(\overrightarrow{AB}=(2,1,1)\) and \(\overrightarrow{AC}=(1,3,2)\). Their cross product is \((1\cdot2-1\cdot3,\ 1\cdot1-2\cdot2,\ 2\cdot3-1\cdot1)=(-1,-3,5)\), with length \(\sqrt{1+9+25}=\sqrt{35}\). The area of the triangle is half the area of the parallelogram: \(\frac12\sqrt{35}\approx2.96\). For the angle, \(\overrightarrow{AB}\cdot\overrightarrow{AC}=2+3+2=7\) and the lengths are \(\sqrt6\) and \(\sqrt{14}\), so \(\cos A=\dfrac7{\sqrt{84}}\approx0.764\) and \(A\approx40.2^\circ\). As a check, \(\sin A=\dfrac{\sqrt{35}}{\sqrt{84}}\approx0.645\) and \(0.764^2+0.645^2\approx1\).
Note: linear systems in the real world
Large linear systems are among the most common calculations performed by computers. Engineers analyse structures with thousands of unknown forces, economists study the flows between sectors of an economy, and computer graphics uses vectors, dot products and cross products for every rendered frame: the cross product gives the normal to a surface, and the dot product with the light direction decides how bright it looks. Gaussian elimination, the method of this chapter, is still at the heart of the software that solves such problems. For very large systems, computers use refined versions that choose the largest available pivot at each step to limit rounding errors, and iterative methods that improve an approximate solution step by step. Understanding small systems by hand is the best preparation for trusting, and checking, the answers that such software produces.
Watch out: typical mistakes
- Getting the middle component of the cross product wrong: it is \(a_3b_1-a_1b_3\).
- Forgetting that \(\mathbf a\times\mathbf b=-\mathbf b\times\mathbf a\).
- Applying a row operation to the coefficients but not to the right-hand side.
- Concluding "no solution" from a zero row \((0\ 0\ 0\mid0)\), which actually signals a free parameter.
Did you know?
The cross product and the dot product were introduced in the 1880s by Josiah Willard Gibbs and Oliver Heaviside to simplify the equations of electromagnetism. Gaussian elimination is named after Carl Friedrich Gauss, who used it around 1810 to compute the orbit of the asteroid Pallas, but the same method appears in the Chinese text The Nine Chapters on the Mathematical Art, written about two thousand years ago.
Otto’s tip
After solving, substitute the answer back into every original equation.
How to use this University vectors and systems of linear equations revision notes
First, read the key points so that you know where the University vectors and systems of linear equations 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 and systems of linear equations 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 and systems of linear equations 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 University vectors and systems of linear equations 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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