Given u = ⟨1, 3⟩, v = ⟨2, 1⟩, and cos(ϴ) = sqrt2/2, where ϴ is the angle between the vectors, what is the scalar projection uv and the dot product u · v?
uv = 1.41 and u*v = 4.47
uv = 2.24 and u · v = 5.00
uv = 2.24 and u · v = 7.07
uv = 7.07 and u · v = 15.81

Answers

Answer 1

Answer:

Step-by-step explanation:

l

Answer 2

Option C corresponds to the vectors scalar projection and dot product, which are uv = 2.24 and u v = 7.07.

To find the scalar projection (uv) and the dot product (u · v) of vectors u and v, use the following formulas:

Scalar Projection (uv) of u onto v:

uv = |u| × cos(ϴ)

Dot Product (u · v) of u and v:

u · v = |u| × |v| × cos(ϴ)

u = ⟨1, 3⟩ and v = ⟨2, 1⟩, and cos(ϴ) = √(2)/2,

calculate the magnitudes of u and v first:

|u| = √(1² + 3²) = √10

|v| = √(2² + 1²) = √5

Now, calculate the scalar projection (uv) and the dot product (u · v):

Scalar Projection (uv) of u onto v:

uv

= |u| × cos(ϴ)

= √10 × √(2)/2

= √(10/2)

= √5 ≈ 2.24

Dot Product (u · v) of u and v:

u · v

= |u| × |v| × cos(ϴ)

= √10 × √5 × √(2)/2

= √(1052)/2

= √100/2

= √50

≈ 7.07

Therefore, the scalar projection and dot product of the vectors is equal to option c. uv = 2.24 and u · v = 7.07.

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