8.2 Affine Independence 47714. (T/F) Given S ={b1.....bx} in R", each p in aff S has aunique represcntation as an affine combination of bi...bk-%3D15. (T/F) If S = {v.....v} is an affinely independent set inR" and if p in R" has a negative barycentric coordinatedetermincd by S, then p is not in aff S.16. (T/F) When color information is specificd at cach vertexV1, V2. V3 of a triangle in R', then the color may be inter-polated at a point p in aff {v, V2, V3} using the barycentriccoordinates of p.17. (T/F) If VỊ, V2, V3. a, and b are in R3 and if a ray a +ib for120 intersects the triangle with vertices v. V2, and v3, thenthe barycentric coordinates of the interscction point arc allnonnegative.pue18. (T/F) If T is a triangle in R and if a point p is on an edgeof the triangle, then the barycentric coordinates of p (for thistriangle) arc not all positive

Note:- As per our guidelines, we can answer the first one as all the are not subpart of a particular problem. Please resubmit the problem with mentioning the part number. Thank you Problem:- 14 Given that S=b1, b2, .....bk in ℝn According to a known result, ∙ For a set S=b1, b2, .....bk in ℝn, any x∈S can be written as uniquely as the affine combination of b1, b2, .....bk for some sequence of scalars in ℝ if and only if the set of points b1, b2, .....bk is affinely independent. Proof:- Let the set of points b1, b2, .....bk is affinely independent but every x∈S can not be written as uniquely as the affine combination of b1, b2, .....bk that is, there exists another sequence of scalars such that x=λ1b1+....... + λkbk =μ1b1+....... + μkbkorx=λ1b1+....... + λkbk ----(1)x=μ1b1+....... + μkbk ----(2) and λi≠μi ; i∈1, 2, ...., kHere,λ1+.......+ λk=1 and μ1+....... + μk=1 subtract (2) from (1), we get then x-x=λ1b1+....... + λkbk - μ1b1+....... + μkbk⇒0=λ1- μ1 b1+....... + λk - μk bk ⇒λ1- μ1= λ2- μ2=......=λk - μk=0 as bi's are non zero i∈1, 2, ...k⇒λ1= μ1, λ2= μ2, ...... , λk = μk So, our assumption is wrong. And, every x∈S can be written as uniquely. Conversely, Let every x∈S can be written as uniquely as the affine combination of b1, b2, .....bk that is, x=λ1b1+....... + λkbk with λ1+.......+ λk=1 And we'll show that b1, b2, .....bk is linearly independent. Let the sequence of scalars is unique then α1b2-b1 + ....... +αkbk-b1 =0 As, we assume combination is unique then b2-b1, ....... ,bk-b1 are linearly independent. And, sequence of scalars is unique. Using above result we can say that the statement is true if and only if b1, b2, .....bk is affinely independent. As, nothing is mentioned about b1, b2, .....bk So, in general the statement is false ( If the b1, b2, .....bk is not affinely independent) Answer:- Statement is False.