Hint 2: A normal vector to the plane ax + by + cz = d is (a, b, c).
Hint 3: The projection of m onto the direction of N is (m, N) / |N| (here (m, N) is the scalar, or dot, product).
Let be the projection of the three-dimensional space onto the plane . Find the matrix of this linear transformation in the standard basis i, j, k.
[Hint: Find the coordinates of the normal vector N to the plane. Then use the fact that the projection of a basis vector m (m = i, j, or k) onto the plane is m - a where a is the projection of m onto the direction of N.]
You may wish to review the material about projections and scalar (dot) product.
I am sorry, hint 3 from post #2 should be as follows.
Hint 3: The projection of m onto the direction of N is ((m, N) / |N|2) N, where (m, N) is the scalar, or dot, product.
Previously I said that the projection was (m, N) / |N|, but this is the scalar projection, not a vector. To get the vector projection onto the direction of N, the scalar projection must be multiplied by the unit vector in the direction of N, which is N / |N|. This article in Wikipedia describes projections well.
(m, N) / |N|2) N = m, N) / 3) N.
Now for m = i = (1,0,0) we have .
Now for m = j = (0,1,0) we have .
Now for m = k = (0,0,1) we have .
so . From here we should do m - a. Do I do (i, j, k) - to give me a 3x3 matrix?
I'll drop the bold font.
((i, N) / |N|²) N is the projection of i onto N (the result is a vector), and similarly for j and k. You subtract that projection from the corresponding original vector (i, j or k), and you get the projection of that vector onto the plane.
I is a basis for U and is a basis for V, A is a linear transformation from U to V,
then the "matrix representation" of A in those two bases requires calculating [tex]A(u_i)[tex] and writing as a linear combination of the vectors. The coefficents of in form the column of matrix.
So what are (1, 0, 0), (0, 1, 0), and (0, 0, 1) projected to?