Vectors and spaces #10 - Q&A

Question 1 determine whether W is a subspace of the vector space. W(z,y):z-y=1}, VR² Answer 1 By the definition of a subspace a nonempty subset W of a vector space is called subspace of V, if W is a vector space under the operations of addition and scalar multiplication defined in V. From the above definition it is clear that a subspace must also be a vector space itself and thus must satisfy the axioms of a vector space. However, the additive identity for R that is, (0,0) is not in the subset W. Hence, all the vector space axiom are not satisfied by W, so it is not a vector space, and hence not a subspace. Therefore, W is not a subspace of R. Question 2 Write formulas for the unit normal and binormal vectors of a smooth space curve r(t) Answer 1 Write formulas for the unit normal and binormal vectors of a smooth space curve r(t) The unit binomial vecto B(t) is a vector that is perpendicular to both T(t) and N(t). T(t) N(t)= B(t)T(t) x N(t) Question 3 Which vectors define the complex number plane? Answer 1 1(1,0) and i = (0,1) The complex number plane is usually considered as a two dimensional vector space over the reals. The two coordinates represent the real and imaginary parts of the complex numbers. As such, the standard orthonormal basis consists of the number 1 and i, 1 being the real unit and i the imaginary unit. We can consider these as vectors (1,0) and (0,1) in in fact, if you start from a knowledge of the real numbers & and want to describe the complex numbers C, then you can define them in terms of pairs of real numbers with arithmetic operations: (a,b)+(c,b)=(a+c,b+d) (this is just addition of vectors) (a,b) (c,b)(ac-bd, ad+bc) The mapping (a,0) embeds the real numbers in the complex numbers, allowing us to consider real numbers as just complex numbers with a zero imaginary part. Note that (a,0) (c,d) (uc, ad) which is effectively scalar multiplication. Question 4 Find the Euclidean distance between u and v and the cosine of the angle between those vectors. State whether that angle is acute, obtuse, or 90, u (-1,-18,0), v = (5, 6, 1, 4) Answer 1 The Eunclidean distance between u and v is the Euclidean norm of the vector uv. Thus, we must find u - v = (- 1, - 1, B, 0) - (5, 6, 1, 4) = (- 6, - 7, 7, - 4) and l(u, v) = ||u - v|| = sqrt((- 6) ^ 2 + (- 7) ^ 2 + 7 ^ 2 + (- 4) ^ 2) = sqrt(150) = 5sqrt(6) Furthermore, the angle 0 between these vectors is given by a * 0 = \langleu, v\rangle/||u - v|| , where (u, v) is a scalar product of u and v. So we compute (u, v rangle= - 5 - 6 + 8 + 0 = - 3 ||u|| = sqrt((- 1) ^ 2 + (- 1) ^ 2 + 8 ^ 2 + 0 ^ 2 = sqrt(66)) ||v||= sqrt 5 ^ 2 + 6 ^ 2 + 1 ^ 2 + 4 ^ 2 = sqrt(78) cos 0 = - 3/(sqrt(66) * sqrt(78)) 0= arccos (-3) 1.613 √66/78 So, this angle is obtuse. Result Distance: d(u, v) = 5sqrt(6) 0= arccos (-3) √6678 it is obtuse 1.613 Question 5 Write an equation of the line that passes through (3, 1) and (0,10) Answer 1 When we are given two points and we have to write the equation that passes through the two points, we can first find the slope. Let's say (3,1) is (3) and (0,10) is (223) Using two points, we can use find the slope with the following formula. (12-1) (22-21) Replace these variables with the values. (10 - 1)/(0 - 3) = - 3 -3 is the slope. Next, we have to find the y-Intercept. We can plug in the values that we know using the following formula: y=mz+b y = - 3x + b Now we can choose one of the points, and plug in the values in the equation above. Let's use (0,10). I = - 3(0) + b b = 10 (Note: The point (0,10) is already on the y-axis, meaning that 10 is the y-intercept. We didn't need to do the step above for this specific problem.) So the complete equation of the line that passes through the two points is: y = - 3x + 10 Answer 2 Step 1: Find the slope (m) The slope of a line passing through two points (xy) and (y) can be calculated using the formula: Substituting the given points, we have m = (1pi - 1)/(2 - 1) = pi/- 3 = - 3 Step 2: Use the point-slope form Now that we have the slope (m = - 3) and one of the points (x_{1}, y_{1}) = (3, 1) we con substitute these values into the point-slope form to get the line: y - 1 = - 3(x - 3) Step 3: Simplify the equation Distribute the -3 on the right side: y - 1 = - 3x + 9 Step 4: Move all terms to one side To get the equation in standard form, we need to move all the terms to one side of the equation 3x + y = 10 This final equation of the line passing through the points (3, 1) and (0,10). Answer 3 The equation of the line passing through (3.1) and (0,10) is: y - 1 = (10 - 1)/(0 - 3) * (x - 3) (12-1) (22-21) Replace these variables with the values. (10 - 1)/(0 - 3) = - 3 -3 is the slope. Next, we have to find the y-Intercept. We can plug in the values that we know using the following formula: y=mz+b y = - 3x + b Now we can choose one of the points, and plug in the values in the equation above. Let's use (0,10). I = - 3(0) + b b = 10 (Note: The point (0,10) is already on the y-axis, meaning that 10 is the y-intercept. We didn't need to do the step above for this specific problem.) So the complete equation of the line that passes through the two points is: y = - 3x + 10 Answer 2 Step 1: Find the slope (m) The slope of a line passing through two points (xy) and (y) can be calculated using the formula: Substituting the given points, we have m = (1pi - 1)/(2 - 1) = pi/- 3 = - 3 Step 2: Use the point-slope form Now that we have the slope (m = - 3) and one of the points (x_{1}, y_{1}) = (3, 1) we con substitute these values into the point-slope form to get the line: y - 1 = - 3(x - 3) Step 3: Simplify the equation Distribute the -3 on the right side: y - 1 = - 3x + 9 Step 4: Move all terms to one side To get the equation in standard form, we need to move all the terms to one side of the equation 3x + y = 10 This final equation of the line passing through the points (3, 1) and (0,10). Answer 3 The equation of the line passing through (3.1) and (0,10) is: y - 1 = (10 - 1)/(0 - 3) * (x - 3)