Determine the domain and range of the given function. The domain is The range is

Mathematics

2 answers:

astra-53 [7]3 years ago

6 0

What is the function?

hoa [83]3 years ago

3 0

Domain: all values that go into a function
range: difference between the lowest and highest values

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Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v, proj,u

11Alexandr11 [23.1K]

Answer:

Take $b = \frac{-17}{25}(7,1)$ as the projection of u onto v and $w \frac{1}{25}(-31,217)$ as the vector such that b+w =u

Step-by-step explanation:

The formula of projection of a vector u onto a vector v is given by

$\frac{u\cdot v}{v\cdot v}v$ , where $cdot$ is the dot product between vectors.

First, let b ve the projection of u onto v. Then

$b = \frac{u\cdot v}{v\cdot v}v= \frac{-6\cdot 7+8\cdot 1}{7\cdot 7+1\cdot 1}(7,1) = \frac{-34}{50}(7,1) = \frac{-17}{25}(7,1)$

We want a vector w, that is orthogonal to b and that b+w = u. From this equation we have that w = u-b = (-6,8)-\frac{-17}{25}(7,1)= \frac{1}{25}(-31,217)[/tex]

By construction, we have that w+b=u. We need to check that they are orthogonal. To do so, the dot product between w and b must be zero. Recall that if we have vectors a,b that are orthogonal then for every non-zero escalar r,k the vector ra and kb are also orthogonal. Then, we can check if w and b are orthogonal by checking if the vectors (7,1) and (-31, 217) are orthogonal.

We have that $(7,1)\cdot(-31,217) = 7\cdot -31 + 217 \cdot 1 = -217+217 =0$ . Then this vectors are orthogonal, and thus, w and b are orthogonal.