Let u1(1/√3,1/√3,-1/√3) , U2=(1/√6, 1/√6, 2/√6) , U3=(1/√2, -1/√2, 0)

Question

A. Show that each U1,U2,U3 is orthogonal to the other two and that each is unit vector
B. Find the projection of E1 on each of U1,U2,U3
C. Find the projection of A=(a1,a2,a3) on U1

Answer 1

A. To show that U1, U2, U3 is orthogonal to each other, we need to show that their dot products are equal to zero.

U1 • U2 = (1/√3)(1/√6) + (1/√3)(1/√6) + (-1/√3)(2/√6) = 1/3 + 1/3 - 2/3 = 0

U1 • U3 = (1/√3)(1/√2) + (1/√3)(-1/√2) + (-1/√3)(0) = 1/√6 - 1/√6 + 0 = 0

U2 • U3 = (1/√6)(1/√2) + (1/√6)(-1/√2) + (2/√6)(0) = 1/√12 - 1/√12 + 0 = 0

Therefore, each of U1, U2, U3 is orthogonal to the other two.

To show that each U1, U2, U3 is a unit vector, we need to show that their magnitudes are equal to 1.

|U1| = √((1/√3)^2 + (1/√3)^2 + (-1/√3)^2) = √(1/3 + 1/3 + 1/3) = √(3/3) = 1

|U2| = √((1/√6)^2 + (1/√6)^2 + (2/√6)^2) = √(1/6 + 1/6 + 4/6) = √(6/6) = 1

|U3| = √((1/√2)^2 + (-1/√2)^2 + 0^2) = √(1/2 + 1/2 + 0) = √(2/2) = 1

Therefore, each of U1, U2, U3 is a unit vector.

B. To find the projection of E1 on each of U1, U2, U3, we can use the formula for vector projection:

proj_U(E1) = (E1 • U) * U

proj_U1(E1) = (E1 • U1) * U1 = (1/√3)(1/√3) + (1/√3)(1/√3) + (-1/√3)(-1/√3) = 1/3 + 1/3 + 1/3 = 1

proj_U2(E1) = (E1 • U2) * U2 = (1/√3)(1/√6) + (1/√3)(1/√6) + (-1/√3)(2/√6) = 1/6 + 1/6 - 2/6 = 0

proj_U3(E1) = (E1 • U3) * U3 = (1/√3)(1/√2) + (1/√3)(-1/√2) + (-1/√3)(0) = 1/√6 - 1/√6 + 0 = 0

Therefore, the projection of E1 on U1 is 1, the projection of E1 on U2 is 0, and the projection of E1 on U3 is 0.

C. To find the projection of A on U1, we can use the same formula as before:

proj_U1(A) = (A • U1) * U1

proj_U1(A) = (a1,a2,a3) • (1/√3,1/√3,-1/√3) * (1/√3,1/√3,-1/√3)

proj_U1(A) = (a1/√3 + a2/√3 - a3/√3) * (1/√3,1/√3,-1/√3)

proj_U1(A) = ((a1 + a2 - a3)/3) * (1/√3,1/√3,-1/√3)

Therefore, the projection of A on U1 is ((a1 + a2 - a3)/3) * (1/√3,1/√3,-1/√3).

Answer 2

close, but the projection of a vector on another is a vector, not a scalar.
You said that the projection of E1 on U1 is 1
but that is not a vector.

Answer 3

You're correct, my apologies for the mistake. Let's correct it.

B. To find the projection of E1 on each of U1, U2, U3, we can use the formula for vector projection:

proj_U(E1) = ((E1 • U) / |U|^2) * U

where • represents the dot product and |U| represents the magnitude of vector U.

proj_U1(E1) = ((E1 • U1) / |U1|^2) * U1 = ((1/√3)(1/√3) + (1/√3)(1/√3) + (-1/√3)(-1/√3)) / ((1/√3)^2 + (1/√3)^2 + (-1/√3)^2) * (1/√3, 1/√3, -1/√3)

Simplifying, we have:
proj_U1(E1) = (1/3 + 1/3 + 1/3) / (1/3 + 1/3 + 1/3) * (1/√3, 1/√3, -1/√3)

proj_U1(E1) = 1 * (1/√3, 1/√3, -1/√3) = (1/√3, 1/√3, -1/√3)

Similarly, we can find the projections for U2 and U3:
proj_U2(E1) = ((E1 • U2) / |U2|^2) * U2 = ((1/√3)(1/√6) + (1/√3)(1/√6) + (-1/√3)(2/√6)) / ((1/√6)^2 + (1/√6)^2 + (2/√6)^2) * (1/√6, 1/√6, 2/√6)

Simplifying, we have:
proj_U2(E1) = (1/6 + 1/6 - 2/6) / (1/6 + 1/6 + 4/6) * (1/√6, 1/√6, 2/√6)

proj_U2(E1) = 0 * (1/√6, 1/√6, 2/√6) = (0, 0, 0)

proj_U3(E1) = ((E1 • U3) / |U3|^2) * U3 = ((1/√3)(1/√2) + (1/√3)(-1/√2) + (-1/√3)(0)) / ((1/√2)^2 + (-1/√2)^2 + 0^2) * (1/√2, -1/√2, 0)

Simplifying, we have:
proj_U3(E1) = (1/√6 - 1/√6 + 0) / (1/2 + 1/2 + 0) * (1/√2, -1/√2, 0)

proj_U3(E1) = 0 * (1/√2, -1/√2, 0) = (0, 0, 0)

Therefore, the projection of E1 on U1 is (1/√3, 1/√3, -1/√3), the projection of E1 on U2 is (0, 0, 0), and the projection of E1 on U3 is (0, 0, 0).

C. To find the projection of A on U1, we can use the same formula as before:

proj_U1(A) = ((A • U1) / |U1|^2) * U1

proj_U1(A) = ((a1,a2,a3) • (1/√3,1/√3,-1/√3)) / ((1/√3)^2 + (1/√3)^2 + (-1/√3)^2) * (1/√3,1/√3,-1/√3)

proj_U1(A) = ((a1/√3 + a2/√3 - a3/√3) / (1/3 + 1/3 + 1/3)) * (1/√3,1/√3,-1/√3)

proj_U1(A) = ((a1/√3 + a2/√3 - a3/√3) / 1) * (1/√3,1/√3,-1/√3)

proj_U1(A) = (a1/√3 + a2/√3 - a3/√3) * (1/√3,1/√3,-1/√3)

Therefore, the projection of A on U1 is (a1/√3 + a2/√3 - a3/√3) * (1/√3,1/√3,-1/√3).