## SUGGESTIONS FOR HOMEWORK #9

Homework suggestions:

5.2
#32(b) Use the fact that when all the entries of a matrix except for the second column are held fixed, the determinant is a linear function of the second column. Use also that

```
3    1          1
5 =  1  +    2  2
7    1          3

```

5.3 #4 (not bracketed) Let u,v,w be the vertices of the triangle. Thus, for example, u = a_1
a_2

The sides of the triangle at the vertex u are given by v-u, w-u, emanating from the tip of u. The area of the triangle is therefore half the absolute value of the determinant of the matrix whose columns are v-u, w-u.

Now look at the 3 by 3 determinant in the book. If you subtract the first column from the each of the other two the third row becomes 1 0 0. Expanding with respect to this row gives 1 det B, where B is the matrix with columns v-u, w-u. Thus, the area of the triangle is also half the determinant of the given 3 by 3 matrix.

#5 (not bracketed) The cone with vertex P over a region R in a plane (P is not in the plane) consists of all points that lie on a line joining P to a point of R. When R is a disk and P is on a line perpendicular to the plane through the center of the disk, this gives an ordinary cone. The volume of a cone is (1/3)(altitude)(area of the base), no matter what shape the base is. The reason is that if the altitude is h and we let x be the distance of a variable point on the altitude to the vertex (x runs from 0 to h) then the area of a cross-section through a point on the altitude at distance x from the vertex is A(x/h)^2. One integrates this as x goes from 0 to h to get the formula Ah/3.

Pick two of the three vectors determining the parallelepiped: they determuine a parallelogram. Say that parallelogram has area A. Say the height of the parallepiped is h. The volume of the parallelepiped is hA. The tetrahedron can be thought of as a cone of height h over a triangle that has half the area of the parallegram, i.e., A/2. Thus, the volume of the tetrahedron is (1/3)h(A/2) = Ah/6, or one sixth the volume of the parallepiped.

#6 (to be turned in) Referring to the analyses above, you should be able to calculate the ratio of the area of the specified triangle to the volume of the specified tetrahedron. The answer is a completely specific real number that does not depend in any way on the values of the a's, b's, and c's.