Question

Let u= ( 5, 7, 8, -6) and v = (8,8,1, -4). Find ∥u−projᵥ u∥. Note: You can partially check your work by first calculating the projection of u onto v and then proceed to find the norm as specified. Please provide your solution.

Answer 1

The question asks for the norm of the vector u minus its projection onto vector v. To solve this, find the dot products needed for the projection, subtract the projection from the original vector u, and then calculate the magnitude of the result.

Step-by-step explanation:

We need to calculate ‖u−projₕ u‖. First, let's find the projection of u onto v (πₕ(ᵔ)).

The projection is given by:

projₕᵔ = ᵔ ⋅ (ᵔ ∙ π) / (π ∙ π) × π
We have:

• ᵔ = (5, 7, 8, -6)
• π = (8, 8, 1, -4)

Calculate the dot product ᵔ ∙ π:

5*8 + 7*8 + 8*1 + (-6)*(-4) = 40 + 56 + 8 + 24 = 128
Calculate the dot product π ∙ π:8*8 + 8*8 + 1*1 + (-4)*(-4) = 64 + 64 + 1 + 16 = 145
Now compute the projection:projₕᵔ = (128/145) × (8, 8, 1, -4)
Then, we find the vector ᵔ - projₕᵔ. Finally, we calculate its norm (magnitude):‖u - projₕ u‖ = √((u1 - projₕ1)^2 + (u2 - projₕ2)^2 + (u3 - projₕ3)^2 + (u4 - projₕ4)^2)
Remember, we need the individual components of the projection to subtract from π and then square and add these components and then take the square root for the magnitude.