Hi, I need help on this linear algebra question. Please use #2 as a reference, I need help solving #4. Thank you.

4. Let's do one more linear transformation. (a) (b) (C) (d) (B) Let S : R3 > R3 be projection onto the plane at +24 + z = 0. Geometrically, this means we mapping every vector to its 'shadow' on the plane. Mathematically, this means that S (u) = u for any vector u in the plane, and S(v) = 0 for any vector v normal to the plane. Using this description, nd S(v1), 3(v2), and S(v3), where v1,v2, v3 are from problem 2. :c If x = y , solve the equation z $1V1 + 332V2 + 333V3 = X. Your answer should involve 56,31, 2. (This should be the same answer as in problem 3; you can just copy it here.) Using your answers to (a) and (b), nd a formula for the linear transformation 8(x). (Hint: if x = clvl + c2v2 +c3v3, then 3(x) = 613(V1) + c23(v2) + 035(v3).) What is the kernel of 8? Is 8 invertible? Explain why or why not. One more, just for fun. Is T o S = S o T? Explain why or why not geometrically, and then check via the formulas for S and T. 2. Let's continue with the plane at: + y + z = O. (a) Find a basis for the plane .7: + y + z = 0. It should contain two vectors {V1, v2}. (b) Find a nonzero vector v3 that is normal to the plane x + y + z = 0. (If you haven't taken calculus where you learn about normal vectors, use this definition: V3 is normal to the plane x + y + z =0 if, for any u in the plane, V3 . u = 0. (c) Show that {v1, V2, V3} is a basis for R3