Download A Linear Algebra I

A
Linear Algebra I
MCS
Darmstadt University of Technology
Department of Mathematics
Winter 2001/02
Prof. Dr. Günter Leugering
Georg Hofmann, Dr. Werner Nickel
11 Dec 2001
Tutorial no. 7
The Cross and the Mixed Product
The problem of finding a vector in R3 which is perpendicular to two given vectors in R3 is a task often
encountered in geometric and physical application of vector calculus. In this tutorial we will study an
operation for vectors which solves this problem. More details can be obtained from Chapter 2.7 of the
lecture notes, which was not treated in the lectures.
T19 Read Definition 2.7.1. and calculate v × w for v := (1, 1, 2)T und w := (2, −1, 3)T .
An important difference between the scalar product and the cross product is that the scalar multiplication
yields a number while the cross product yields a vector. The following exercise combines the two products
and shows that v × w is perpendicular to v and w.
T20 Prove the following identities for vectors u, v and w in R3 :
a) hu, u × vi = 0, i.e. u × v is perpendicular to u;
b) hv, u × vi = 0, i.e. u × v is perpendicular to v;
c) ||u × v||2 = ||u||2 ||v||2 − hu, vi2 (Lagrange identity);
d) u × v = −v × u (anti-commutativity).
The cross product has a useful geometric interpretation which we will now work out.
T21
a) Let u and v be vectors in R3 and let θ be the angle between them. Then we have ||u × v|| =
||u|| · ||v|| · sin θ.
b) If u and v are vectors in R3 , then ||u × v|| is the area of the parallelogram spanned by those
vectors.
The next two exercises apply what we have seen so far.
T22
a) Compute the area of the triangle with vertices P1 := (1, −1, 2)T , P2 := (7, 4, 3)T , P3 := (5, 2, 1)T .
b) Is there a number t ∈ R such that the vector (1, t, 6)T is perpendicular to the vectors (1, 1, −1)T
and (2, −4, 3)T ?
c) Let A := (1, 1, 1)T and B := (0, −1, 1)T be points in R3 and E the plane which is determined by
the origin and A and B. Determine a point C on the line through the origin q
which is perpendicular
to the plane E such that the area of the triangle with vertices A, B, C is
63
2 .
Now the mixed product can be defined using the cross product.
T23 Read Definition 2.7.6. and compute the mixed product of the vectors u := (1, 4, −4)T , v := (0, 3, 2)T ,
w := (3, −2, −5)T .
The following exercise illustrates why the mixed product is also called parallelepipedal product. The
modulus of the mixed product is the volume of the parallelepipedon spanned by the vectors involved.
T24 Show that the volume of the parallelepipedon spanned by the vectors u, v, w in R3 is |hu × v, wi|.