But we'll study these definitions in this assignment. Your goal is to show there are two planes in R4 that intersect in a point, that is, the origin, that is, the vector ERA. To do the assigned problem in Part 2, recall the definition of a line in Rn. Definition 1. Suppose u E R" is a non-zero vector. Then the line through u is the set Lu = {v | v = ) . u for some scalar ) ER) . Here are some examples of lines in 2. Example 1. Let e1 = Notice the E R2. Notice that v / 0. Then the x-axis in R2 is the set Lei - {. el |ER} = 0 NERThe y-axis is the set Lez def { . v | AER} = ooyo JER, Example 2. Let u = OOHO E R4. Then the y-axis in R4 is the set Lu -{ . u | ER} = coyo ER Theorem 1. There are two lines in R2 whose intersection is a single point, the origin. That is the intersection of the two sets representing these two lines is the set containing just he origin in R, o} = {}.Proof. Consider the line X determined by the vector _ def l 81 = [0]. was was] Let Y be the line in R2 determined by the vector _ def [0] 82: 1 . mores- My goal is to show that the intersection of these two sets, that is the set of all vectors that are both in X and in Y, is the set 0 0 Note. The set of all vectors that are in X and in Y, that is the intersection of the lines X and Y, is often denote X D Y. That is, aER}. That is, Note. The set of all vectors that are in X and in Y, that is the intersection of the lines X and Y, is often denote X ('1 Y. 8] 6 X H Y. In fact [3] = 0- [a] E X. Also, the vector [ BMW 0 Observe rst that the vector [ 0 161/ since So [3] is in both lines X and Y (I: Now notice that if a vector [y] 6 X then [an] = [A] for some value of A , and so, y = 0. y 0 Also, if a vector [i] 6 Y then [m] = [A] for some value of m and so a: = 0. {B This means that if [y] e X D Y then x = 0 and y = 0. So [m] = [0]. It follows that ma} 9 0 as claimed. That is the intersection of the lines X and Y contains a single vector, the origin in R2. D To work today's assignment, we need to remind you of the denition of a plane. To do so we must rst dene non-colinear vectors in R". Definition 2. Two vectors u and v in R" are non-colinear, if neither u or v are equal to 0, and if u # 1 . v for any ). Example 3. Let u = OOOH , E R4. Let v = OOHO E R4. Then u and v are non-colinear. To see this, note first that neither u or v is the zero vector since both vectors have a non-zero coordinate. If u and v are colinear then u = 1 . v for some number 1. But, u = OOOH and 1 . v = 1 . OOHO for some number 1. If these vectors are equal for some choice of 1 then their coordinates must be the same. But the first coordinate of u is 1, and the first coordinate of A . v is 0. Since 1 / 0, u and v must be non-colinear.Now recall the definition of a plane in Rn Definition 3. Suppose u and v are non-colinear vectors in R". Then the plane containing u and v is defined to be Pu,u def {a . u + B . va, BER}.To get you started on the hand-in assignment, I'll define a plane in R4 for you. Let * = This is a non-zero vector in R4. Let v = OOHO LOOO- be another non-zero vector in R4. Note that u * 1 . v for any number ), since uf d . v= boyo for any choice of 1. By the definition of non-colinear vectors, that is, since both u and v are non-zero and since u * *v for any choice of scalar 1, the vectors u and v are non-colinear. So by the definition of a plane, Pu, is a plane. In fact, Pu,v = {a . u + B . v} = a, BERProblem: Define another plane that you believe with intersect the plane I just defined at only one point.Extra credit problem: (For up to 5 points extra credit.) Write a clear and concise argument that the intersection of the plane which I described above, P55, with the plane you just dened, is a set containing a single vector, the vector 5, and you may receive an additional 5 points of extra credit