Vectors and spaces #8 - Q&A

Question 1 Find the area of the parallelogram determined by the given vectors u and L. u = (L - 1, 2) v = (0,3,1) Answer 1 av = (1, - 1, 2)(0, 3, 1) =([ matrix -1&2\\ 3&1 matrix ],-| matrix 1&2\\ 0&1 matrix |*| matrix 1&-1\\ 0&3 matrix ])=(-7,-1,3) So ||u* v||=||-7-1 3||= sqrt((- 7) ^ 2 + (- 1) ^ 2 + 3 ^ 2) = sqrt(49 + 1 + 9) = sqrt(59) Answer Question 2 (a) Find the point at which the given lines intersect (x, y, z) = 1 (b) Find an equation of the plane that contains these lines. Answer 1 Set the two r vectors equal to each other. r = (2, 3, 0) + t(3, - 3, 3) = (2 + 3t, 3 - 3t, 3t) r = (5, 0, 3) + s(- 3, 3, 0) = (5 - 3s, 3s, 3) 2 + 3t = 5 - 3s t - 3t = 3s 3t = 3 can be easily verified that t = 1 and a = 0 divide both sides of the last equation. by 3 and substitute the result into any of the two above it). Then this immediately tells us t(2 + 3t, 3 - 3t, 3t) = (5, 0, 3) = (5 - 3s, 3s, 3) is the point of intersection. b) Now, the normal vector of the plane is perpendicular to the directional vectors of both of the lines contained in the plane. Take the cross product of the two directional vectors to obtain this normal vector: < 3 > x < - 3, 3, 0 ,-9,0>. Now, any nonzero scalar multiple of this normal vector is also a normal vector. Thus, < 1, 0 > k normal vector. With the normal vector and a point on the plane, such as the point of intersection, (5,10,3), we can write the equation of the plane containing the two given lines. 1(x - 5) + 1(y) + 0(x - 3) = 0 which simplifies to x - 5 + y = 0, epsilon x + y = 5 Question 3 Describe the vector space's zero vector (the additive identity). Answer 1 So, (1.4) Find the vector that has following property: (4) Find the vector that has following property: ()+=(1+x+2+3+4) () + From properties of addition in R, we know that xxx0 So, additive identity vector is x (0,0,0,0). Result: (0,0,0,0) Answer 2 In the vector space R³, the zero vector, denoted as o, is a special vector that plays a crucial role in vector arithmetic. It has the property that when added to any vector ve R, it does not change the vector. More formally, for any vector v+0 As shown above, adding the zero vector to any vector in results in the original vector itself. Geometrically, the zero vector represents the origin of the coordinate system in z. To further understand the properties of the zero vector, let's consider some operations involving it. Firstly, we can subtract a vector from itself: Subtracting a vector from itself results in the zero vector. Secondly, we can multiply the zero vector by any scalar value 0 Multiplying the zero vector by any scalar yields the zero vector itself. These properties showcase the unique role of the zero vector in vector spaces. Furthermore, the zero vector has another important property related to vector addition. Consider two vectors u and v in st. The sum of u and if is given by: If we introduce the negative of a vector, denoted as-v, which is obtained by multiplying by-1, we can rewrite the sum of u and v as follows: +23+0 +++ (+(-))+1 (42+(2)+v )2-) + (( + (+(-))+ +++ ()+ (-)+ (v)+v= This property implies that the negative of a vector cancels out the original vector when added together, resulting in the zero vector. In summary, the zero vector in R¹ is a special vector that does not change other vectors when added to them. It represents the origin of the coordinate system in. (2)+ 0 (-)+ We can observe that (v) + v results in the zero vector Answer 3 Step 1 o is an element of 2: ER. Step 2. Addition with a leaves a vector unchanged. For any vector , we have v+0 In other words, adding to any vector does not change its value. Step 3. Scalar multiplication with O gives the zero vector: For any scalar e Rand vector vel, we have c00. Multiplying the zero vector by any scalar results in the zero vector again These properties make o an essential element in vector space operations and ensure that R is a vector space. Answer 4 Here, the o represents the zero vector, and the notation indicates a column vector. The vector has four rows, where each row corresponds to one of the components of the vector. In this case, all the components are zero, so we have four zeros in the column vector. The zero vector 0 is the additive identity in the vector space R³, which means that when it is added to any vector in a³, it leaves the vector unchanged. For example, if we have a vector in R, adding the zero vector to v yields the same vector: +0 r_{4} + 0 Thus, the zero vector o in vector when added to it. the vector that does not change the value of any other Question 4 Find the coordinate vector of w relative to the basis S={1,2} for Ru₁ = (1,-1), u_{2} = (1, 1) w = (1,0) Answer 1 Consider the basis 2S = [u_{1}, u_{2}] * f where epsilon_{1} = (1, - 1), u_{2} = (1, 1) * z v = (1, 0) In order to find the coordinate vector of w relative to the basis 5 we must find scalars a and b such that av buy w Sinco b*v_{2} = a(1, - 1) + b(1, 1) = (a, - a) + (b, b) = (a + b, - a + b) and omega = (1, 0) We must identify the scalars a and b that (a + b, - a + b) = (1, 0) it brings up the linear system a + b = 1 a + b = 10 +o gives us that a+ therefore t = 1/2 +o gives us that ab and hence +61+0 which gives us that 2b = 1 and a = 1/2 Hence (w)=() Question 5 A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win 1.10; if they are different colors, then you win -110; if they are different colors, then you win-1.00. (This is, you lose $1.00.) Calculate (a) the expected value of the amount you win; (b) the variance of the amount you win. Answer 1 We have that xe (0,1,2) and by the nature of the problem, we have that X has Hypergeometric distribution, ie. P(X = i) = (binomial(5,1) * binomial(5,2 - 1))/(binomial(10,2)) Y=1.1* chi (x \in \{0, 2\}) -1* chi (x=1) \ expectation E(Y)=1.18(X(xe (0.2)))-(()) =1.1P(X=0) x in + 1.1P(X = 2) - P(X = 1) 2-1.1-45-45-15 Answer 2 (b) Use that ar(Y) = E(Y ^ 2) - E(Y) ^ 2 Observe that y^ 2 =1.1^ 2 * x (x \in \{0, 2\}) +1^ 2 * x (x=1) mathcal E ( mathfrak r ^ 2 )=1.21P(X= emptyset)+ 1.21P(X = 2) + P(X = 1) = 82/75 82 1 49 Var(Y)=75 225 45