## Tuesday, January 10, 2012

### Linear Algebra for Solid Geometry

We now look at linear algebra applied to solid geometry. In the last set of solutions we saw the clue on how to proceed, which is to find the determinant of a 3 x 3 matrix, viz.

(u1....u2......u3)
(v1....v2.....v3
(w1....w2....w3)

And we saw that the solution was:

Det M = u1[v2 w3 - v3 w2] - u2 [v1 w3 - v3 w1]

+ u3 [v1 w2 - v2 w1]

Now we will apply this to the case of finding the volume of the parallelopiped shown which has 3 sides spanned by the vectors, u, v and w where:

u = (1, 1, 3)

v = (1, 2, -1)

w = (1, 4, 1)

Now, we let the volume be Vol(u, v, w) and:

Vol (u, v, w) = Det [u, v, w]

So that:

(u1....u2......u3)
(v1....v2.....v3
(w1....w2....w3) =

(1.....1......3)
(1....2.....-1)
(1....4........1)

And Det [u, v, w] =

(2....-1)
{4.....1) -

(1....-1)
(1.....1) +

3 (1....2)
(1.....4)

=

[2 -(-4) - (1 - (-1)) + 3(4 - 2)] = 6 - (2) + 3(2)

And: Vol (u, v. w) = 12 - 2 = 10 cubic units

Problems:

1) For a similar solid to that shown, but with vectors:

u = (1, -1, 4)

v = (1, 1, 0)

w = (-1, 2, 5)

Find: Vol (u, v, w)

2) Show that for the spanning vectors:

u = (-2, 2, 1)

v = (0, 1, 0)

w = (-4, 3, 2)

Vol (u, v, w) = 0