9. For the following vectors, x and y, calculate the indicated similarity or distance measures. a. (a) x : (0,0,1,1), y : (2,2,2,2) cosine, correlation, Euclidean b. (b) x : (0,1,0,1), y : (0,1,0,1) cosine, correlation, Euclidean, Jaccard c. (c) x : (1,1,0,1), y : (-1,0,-1,0) cosine, correlation, Euclidean d. (d) x : (1,0,0,1,0,1), y : (0,1,1,0,0,1) cosine, correlation, Jaccard e. (e) x : (2,1,0,2,0,3), y : (1,1,1,0,0,1) cosine, correlation

The answer to this question can be defined as follows:

a)

cosine: 0.707

correlation: 0

Euclidean: 3.16

b)

cosine: 1

correlation: 1

Euclidean: 0

Jaccard: 1

c)

cosine: -0.4082

correlation: 0.5774

Euclidean: 2.64

d)

cosine: -0.333

correlation:  -0.3333

Jaccard: 0.5

e)

cosine: -0.7075

correlation: 0.2132

Step-by-step explanation:

a) cosine:

correlation:

Euclidean:

by applying above formula we get all the values.

b)

cosine: 1

correlation: 1

Euclidean: 0

Jaccard: 1

c)

cosine: -0.4082

correlation: 0.5774

Euclidean: 2.64

d)

cosine: -0.333

correlation: -0.3333

Jaccard: 0.5

e)

cosine: -0.7075

correlation: 0.2132

a is 8.b is 10.c is 10.5. d is -.5

step-by-step explanation:

just minus the smallest number from the biggest.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)