Hi, alison28-ga:
(1) The angle A between two linearly independent vectors in R^3 is
easily determined by the formula for the dot product in terms of the
vectors lengths and the cosine of the included angle:
<u,v> = cos(A) * ||u|| * ||v||
Taking the cube to have unit-length edges parallel to the coordinate
axes, we can position the cube in the "positive" octant (first
orthant), so that the coordinates of the eight vertex are all the
possible triples than can be formed from entries in {0,1}.
That is, the vector from the origin to the opposite corner (1,1,1)
gives as a diagonal of the cube:
u = [1 1 1]
Here I'm making a notational distinction between the row-vector [1 1
1] and the two points in the three dimensional Euclidean space (0,0,0)
and (1,1,1) which define the origin and terminus of that vector.
Likewise one of the three "face" diagonals that meets the "long"
diagonal at the origin is:
v = [1 1 0]
Solving the relation above and using dot product <u,v> and the lengths
||u|| and ||v|| found by the usual "distance" formula:
<u,v>
cos(A) = ---------------
||u||*||v||
= 2 / (SQRT(3)*SQRT(2)) = SQRT(2/3)
We then deduce that A = arccos( SQRT(2/3) ) = 0.84106867... radians or
48.1896851... degrees.
(2) The computation of the "transition matrix" P which converts
coordinates with respect to the ordered basis:
B = { (1,3,4), (2,-5,2), (-4,2,-6) }
into coordinates for the same point with respect to the ordered basis:
B'= { (1,2,-2), (4,1,-4), (-2,5,8) }
can mostly easily be conceived of in two stages. The first and easy
stage is the conversion from B-coordinates into "standard"
coordinates, that is coordinates with respect to the standard basis:
E = { e_1 = (1,0,0). e_2 = (0,1,0), e_3 = (0,0,1) }
which we will show is accomplished by multiplying by a matrix M. The
second stage is the conversion from standard coordinates to
B'-coordinates, also accomplished by matrix multiplication, and the
appropriate product of the two matrices then gives the transition
matrix P which is the first objective in this problem.
A brief reflection of what it means for a point x to have
B-coordinates of (a,b,c) tells us that the point x is a linear
combination of the standard basis vectors as follows:
x = a*(1,3,4) + b*(2,-5,2) + c*(-4,2,-6)
= (1a+2b-4c)*e_1 + (3a-5b+2c)*e_2 + (4a+2b-6c)*e_3
In particular the B-coordinates (1,0,0) correspond to standard
coordinates (1,3,4), etc.
Although it is essentially an arbitrary decision, the coordinates with
respect to an ordered basis are often represented by a column vector.
With this convention the transition matrix M that converts from
B-coordinates to standard coordinates turns out to have three columns,
each of which has entries transposed from the corresponding basis
vector's entries:
/ 1 2 -4 \
M = | 3 -5 2 |
\ 4 2 -6 /
With this convention of column-vectors for coordinates, one then has
the conversion from B-coordinates to standard coordinates represented
as follows:
/ 1a+2b-4c \ / a \
coord. wrt E | 3a-5b+2c | = M * | b |
\ 4a+2b-6c / \ c /
The point here is that M can be written down by inspection. Similarly
the matrix that converts B'-coordinates to standard coordinates could
also be written down by inspection:
/ 1 4 -2 \
G = | 2 1 5 |
\-2 -4 8 /
However we need to convert in the reverse direction, from standard
coordinates to B'-coordinates in order to accomplish the overall
purpose for matrix P, and thus we need to find the multiplicative
inverse of G so that we may set:
P = G^-1 M
One rather convenient way to do this is by constructing a "super-augmented" matrix:
/ 1 4 -2 | 1 2 -4 \
[ G | M ] = | 2 1 5 | 3 -5 2 |
\-2 -4 8 | 4 2 -6 /
then applying the "Swiss Army knife" of numerical linear algebra,
Gaussian elimination/reduction to reduced row echelon form, we find
after the appropriate sequence of elementary row operations that the
reduced row echelon form of this is [ I | P ]. I shall leave this for
the moment as a calculation for the reader to try his or her hand at.
The calculation of Q is similar to that of P, except that we would
apply Gaussian elimination instead to the extended matrix [ M | G ]
and obtain from it [ I | Q ].
One then expects and can easily verify that P and Q are multiplicative
inverses, as multiplying by Q is naturally undoing what multiplying by
P does.
Finally we are asked to determine what B-coordinates [x]_B correspond
to the point x whose B'-coordinates are given by [x]_B' = (-1,0,2).
For this it suffices to multiply Q times the column (-1,0,2)'. [Here
the prime ' denotes transpose, giving a column in place of "row"
(-1,0,2).] If we had not already at this point found Q (as well as
P), then a shortcut calculation which finds only [x]_B can be arranged
by reducing the augmented matrix:
[ M | (-5,8,18)' ]
whose right-hand side (-5,8,18) is G * (-1,0,2)'. Thus to solve:
/ 1 2 -4 | -5 \
| 3 -5 2 | 8 |
\ 4 2 -6 | 18 /
I will return shortly with the various step by step calculations, but
the answer to part (d) is that the B-coordinates of x work out to:
[x]_B = ( 14 11/13, 11 8/13, 10 10/13 )
Possibly these fractional values indicate that the problem was not
copied correctly (or possibly someone simply picked fairly challenging
numbers for hand calculation!), but in any case I thought it most
valuable to sketch out the approach for all parts of this second
question before diving into the details of the computations.
regards, mathtalk-ga |
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