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Linear Algebra · Book or PDF
A First Course in Linear Algebra
A.G. Hamilton · Study material generated by AI from this edition. Not affiliated with the publisher.
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A.G. Hamilton · Study material generated by AI from this edition. Not affiliated with the publisher.
Chapter 10 · pp. 114–131
This chapter uses vectors to describe lines and planes in three dimensions. A line is built from a point and a direction. A plane is built from a point and a vector at right angles to it. The dot product gives a way to measure angles, and solving linear equations shows how lines and planes meet.
A line in 3D is fixed by two points on it, or by one point and a direction. Let A have position vector and let P be any point on the line. Then . The vector points along the line, so it is a multiple of the direction vector. This gives the direction form (pp. 114, 116).
If the line passes through A and B, the direction is . The two-point form is therefore . It is just the direction form with (pp. 114, 116).
Example. For A(2,3,-1) and B(3,1,3), the direction is . The line is (p. 116).
The number is a parameter. As takes every real value, the right-hand side gives the position vectors of all the points on the line. Writing the vector equation one coordinate at a time gives the parametric equations (pp. 115–116).
For the line above:
Putting gives (3,1,3), which is point B. Putting gives A.
The components of the direction vector are called direction ratios. Any vector pointing the same way will do (p. 116).
A common mistake is to think a line has just one correct set of parametric equations. In fact the starting point can be any point on the line. The direction can be any non-zero multiple of the direction vector, including a unit vector. The equations look different, but they produce the same points (pp. 115–116).
Example. Take the line through (-2,5,1) with direction . Using gives the point (-1,4,3). Now use the unit vector instead. Then gives the same point (-1,4,3). The two sets of equations describe the same line, because the parameter values simply rescale.
Two lines in 3D need not meet. To test, give the lines different parameter letters, such as and . Then set the , and coordinates equal (pp. 117–118).
This gives three equations in two unknowns, which is often inconsistent. If row reduction shows the equations are consistent with a unique solution, the lines meet at one point. Substitute the parameter value into either line to find the point.
Example. Line 1 is . Line 2 is . Row reduction gives and . Line 1 at gives (1,3,4). Line 2 at also gives (1,3,4), so the lines meet there (pp. 117–118).
For and , the dot product is
The answer is a single number, not a vector, so it is also called the scalar product. For example, (p. 118).
The angle between non-zero vectors and is the angle AOB, where and . It lies between 0 and . The key rule is
The proof applies the cosine rule to triangle OAB. It then expands in coordinates, and the squared terms cancel (pp. 117–119).
A negative dot product means , so the angle is obtuse.
Example. Let and . Then , and . So and .
Two non-zero vectors are perpendicular (orthogonal) exactly when . This is because , and the reverse also holds for non-zero vectors (p. 120).
Neither vector has to be zero for the dot product to be zero. For example, , so these vectors are perpendicular.
These rules can be checked by working in components (pp. 119–120):
A plane is fixed by a point A on it and a normal vector perpendicular to it. A point P lies on the plane exactly when is perpendicular to . That gives
This is a vector equation but not a parametric one (pp. 121–122).
Example. Take A(3,1,-2) and . Then , which simplifies to (p. 121).
Every plane has an equation , and the vector is a normal. To see why, take two points on the plane. Subtracting their equations shows that the vector between them has dot product 0 with (pp. 121–122).
The angle between two planes is defined as the angle between their normals. Looking edge-on along the line where the planes meet shows that the two angles agree. So
Planes are parallel when their normals are multiples of each other (pp. 123–124).
Example. The planes and have normals and . The second is twice the first, so the planes are parallel. Solving the two equations together gives no solution, so they share no points (p. 124).
Finding common points of planes means solving linear equations. The type of solution shows the geometry (pp. 124–126).
Two planes. If they are not parallel, they meet in a line. With two equations and three unknowns, one variable becomes the parameter. For and , put . Then and , which are parametric equations of the line (p. 125).
Three planes. There are three possible outcomes:
The perpendicular distance from to the plane is
The proof draws the line through P in the normal direction . It finds the parameter where this line meets the plane, which gives the point M. Then it computes (pp. 126–127).
Example. The distance from the origin to is .