Vectors and spaces #9 - Q&A

Question 1 How to calculate the intersection of two planes? These are the planes and the result is gonna be a line in p³ x + 2y + z - 1 = 0 2x + 3y - 2z + 2 = 0 Answer 1 You need to solve the two equations x + 2y + z - 1 = 0 - 2z + 2 = 0 Notice that, these are two equations in three variables, so you have a free variable say then we have z + 2y = 1 - t 2x + 3y = 2t - 2 Solving the last system gives \{x = - 7 + 7t, y = 4 - 4t\} Then the parametrized equation of the line is given by (x, y, z) = (- 7 + 7t, 4 - 4t, t) = (- 7, 4, 0) + (7, - 4, 1) * t Answer 2 The first plane has normal vector 1 (2) 1 and the second has normal vector 2 (3) so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors, ist, so the direction vector of the line of intersection is 1 2 -7 (2)(3) = (4) 1 Next, we need to find a particular point on the line. Here, since a line that isn't parallel to any coordinate plane passes through all three, you can check if it is parallel to one by using the directional vector. Since here, the line passes through all three planes, we can try y=0(since it passes through the x-z plane) and solve the resulting system of linear equations: x + z - 1 = 0 2x - 2z + 2 = 0 giving epsilon = 0, z = 1 thus the line of intersection is -7t {(4t): teR} 1-t Question 2 Find the distance from the origin to the line x = 1 + t y = 2 - t z = - 1 + 21 Answer 1 x = 1 + t, y = 2 - t, z = - 1 + 2t The line has direction vector <1,-1, 2> We can make a vector from the origin to a point on the line by <1+1 2-1-1+2t> This vector has minimum magnitude when it is perpendicular to the lines Question 3 How to find absolute value of a vector in this question? a = \{t, 3, 5\} Then in formulas see this lal Answer 1 Absolute value of a vector means taking second norm of the vector Le. x. That means the same thing as I don't understand why some top researchers in computer science abuse the notation where ixl is widely used for absolute value of scalars in math. I have also seen papers using (xi to mean absolute value of each component of the vector x. In summary, meaning will depends upon the context and author. Answer 2 Keep in mind that absolute value is distance from zero. So you can use the distance formula to find the absolute value: sqrt(x ^ 2 + y ^ 2 + z ^ 2) Question 4 Find the distance between two planes: C_{1} / x + y + 2z = 4 and C_{2} / 3 * x + 3y + 6z = 18 And find the other plane C3 C₁ that has the distance d po the plain Ca Answer 1 Well, we can rewrite C_{2} / 3 * x + 3y + 6z = 18 Two plain have normal n^ = [3, 3, 6] Chose P, Q from C_{t}, C respectively. Say, P(5, 0, 0) and Q(- 1/2, 0, 0) cup QP^ = [- 11/2, 0, 0] Then Proj{QP= QPn = - 11/36 * [3, 3, 6] = [- 11/12, - 11/12, - 11/6] (3,3,6) 9+9+36 Thus, the distance its ||Proj\ n ^ \ QP^ ||= sqrt((11/12) ^ 2 + (11/12) ^ 2 + (11/6) ^ 2) = sqrt(121/24) Answer 2 For a plane defined by ax + b_{1} + cz = d the normal is said to be (a,b,c). This is a direction so we can normalise It * (1, 1, 2)/(sqrt(1 + 1 + 4)) = (3, 3, 6)/(sqrt(9 + 9 + 36)) parallel and we can write the normal as 1/(sqrt(6)) * (1, 1, 2) which means these two planes are Now we should find two points on the planes Let y = 0 and 20, and find the corresponding z volues. For c_{v} 4 and for C_{21} z = emptyset, So we know C_{1}' contains the point (4,0,0) and C_{2} contains the point t(6, 0, 0) The distance between these points is 2 and the direction is (1,0,0). that this is not the shortest distance between these two points as (1,0,0) ne 1/(sqrt(6)) * (1, 1, 2) ; the direction is not perpendicular to these planes. But this is ok as we dot product between (1,0,0) and is perpendicular to the planes. (1, 0, 0) * 1/(sqrt(6)) * (1, 1, 2) = 1/(sqrt(6)) 1/(sqrt(6)) * (1, 1, 2) * t work out the proportion of the distance that So the distance between the two planes is 2/(sqrt(6)) The last part is to find the plane which is the same distance away from C_{2} C_{1} but in the opposite direction. We know the normal must be the same 1/(sqrt(6)) * (1, 1, 2) . Using this we can write C_{3} / x + y + 2z = a and determine a. When t = 0, z = 0w moved from 4, 0, 0 ) (6,0,0 move the same distance again (6, 0, 0) (8,0,0) subset (8, 0, 0) * ison*C_{3} Therefore, a = 8 So final equation of the plane is: C_{3} / x + y + 2z = 8 Question 5 Find out if the set that can perform the specified operations is a vector space. Identify the vector space axioms that are false for those that are not vector spaces. the collection of all real numbers with addition and multiplication operations. V is not a vector space, and Axioms 7,8,9 fail to hold. V is not a vector space, and Axiom 6 fails to hold. V is a vector space. V is not a vector space, and Axiom 10 fails to hold. V is not a vector space, and Axioms 6-10 fail to hold. Answer 2 Answer: -In Option LV is not a vector space, and Axioms 7, 8, and 9 fail to hold. -In Option 2. Axiom 6 fails to hold, but the specific violation is not specified. -In Option 3, V is a vector space. In Option 4, V is not a vector space, and Axiom 10 fails to hold. -In Option 5, Axioms 6-10 fail to hold, but the specific violations are not provided. Explanation: To determine if the set of all real numbers, denoted as R, with addition and multiplication operations forms a vector space, we need to check if all the vector space axioms hold. Let's examine each option and identify which axioms fail in the cases where the set is not a vector space. Option is not a vector space, and Axioms 7, 8, and 9 fail to hold. The vector space axioms are as follows: Axiom 1: Closure under addition. Axiom 2: Associativity of addition Axiom 3: Existence of additive identity. Axiom 4: Existence of additive inverse. Axiom 5: Closure under scalar multiplication, Axiom 8: Associativity of scalar multiplication. Axiom 7: Distributivity of scalar sums. Axiom 8: Distributivity of vector sums. Axiom 9: Compatibility of scalar multiplication with field multiplication. Axiom 10: Identity element of scalar multiplication properties. Let's go through them: To verity Axioms 7, 8, and 9, we need to check if the specified operations satisfy the given Axiom 7: Distributivity of scalar sums VabERVIEV(a+b)xax + bx Consider 2.3.4. Then we have: (2+3)-4-5-4-20 2-4-3-4-8+12-20 The result holds, so Axiom 7 is satisfied. Axiom 8: Distributivity of vector sums Va e RVs, ye Via(x+y)= ax + ay Considera 2,x3,y4,. Then we have: 2-(3+4)-2-7-14 2-3+2-4-6+8-14 The result holds, so Axiom 8 is satisfied. Axiom 9: Compatibility of scalar multiplication with field multiplication Va,b E R. Vx EV:(ab)x = a(bx) Consider 2,63,x4. Then we have: (2-3)-46-4-24 2-(3-4)2-12-24 The result holds, so Axiom 9 is satisfied. Therefore, in Option 1, the set i is not a vector space, and Axioms 7, 8, and 9 fail to hold. Let's continue analyzing the remaining options. Option 2:1 is not a vector space, and Axiom 6 fails to hold. In this case, Axiom 6, which states associativity of scalar multiplication, is violated. However, it is not specified how Axiom 6 fails. Since the information provided is insufficient, we cannot determine the specific condition under which Axiom 6 is violated, Option 3 V is a vector space. in this option, it is stated that
is a vector space. Therefore, all the vector space axioms hold for the set of real numbers with the specified operations. Option 4: V is not a vector space, and Axiom 10 falls to hold. Axiom 10, which refers to the identity element of scalar multiplication, is violated in this case. However, it is not specified how Axiom 10 fails. Without further information, we cannot determine the exact condition that causes Axiom 10 to be violated. Option 5: 1 is not a vector space, and Axioms 6-10 fail to hold. In this option, it is stated that Axioms 6-10 fail to hold. Similar to Option 2, the specific details of which axioms fail and under what conditions are not provided. Thus, we cannot determine the exact nature of the violations. Answer 3 The vector space axioms are as follows: 1. Closure under addition: Vu,veVu+ver 3. Commutativity of addition: Vu, ve Vu+v=v+u 4. Identity element of addition: 30 € v such that vu € Vu+0= 2. Associativity of addition: vuv.we V. (u+v)+wa+(vw). 5. Inverse elements of addition: vu e V, there exists - such that a+()=0 6. Closure under scalar multiplication: vu ev and vee R, cu eV. 7. Associativity of scalar multiplication: vu EV and ve, de BR. (ed)u=c(du). 8. Distributivity of scalar sums: Vu eV and ved R. (c+d)ucu + du 9. Distributivity of vector sums: vu, ve V and ve E R. c(u+v) 10. Identity element of scalar multiplication: Vu V, 1uu. cu Now let's evaluate each given statement and identify the vector space axioms that are false: 1. V is not a vector space, and Axioms 7, 8, and 9 fail to hold. 2. 1 is not a vector space, and Axiom & fails to hold 3. V is a vector space. 4. 1 is not a vector space, and Axiom 10 fails to hold. 5. V is not a vector space, and Axioms 6-10 fail to hold. Therefore, the correct answer is statement 3: V is a vector space.