This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here are the solutions to your questions: 1. Define the following and give five examples for each: Collinear Vectors: Vectors are collinear if they lie on the same line or are parallel to each other. This means one vector can be expressed as a scalar multiple of the other (u = kv for some scalar k). Examples: 1. u = i, v = 3i 2. u = 2j, v = -6j 3. u = i + j, v = 2i + 2j 4. u = (1, 2, 3), v = (3, 6, 9) 5. u = -k, v = 4k Non-collinear Vectors: Vectors are non-collinear if they do not lie on the same line and are not parallel. One vector cannot be expressed as a scalar multiple of the other. Examples: 1. u = i, v = j 2. u = i + j, v = i - j 3. u = (1, 0, 0), v = (0, 1, 0) 4. u = (1, 2, 0), v = (1, 0, 2) 5. u = i + j + k, v = i + 2j + 3k 2. Define the following and give five examples for each: Coplanar Vectors: Vectors are coplanar if they lie in the same plane. For three vectors, they are coplanar if their scalar triple product is zero (a · (b × c) = 0). Any two vectors are always coplanar. Examples: 1. u = i, v = j, w = i + j (all in the xy-plane) 2. u = (1, 0, 0), v = (0, 1, 0), w = (2, 3, 0) (all in the xy-plane) 3. u = i, v = 2i, w = 3i (all collinear, hence coplanar) 4. u = (1, 1, 1), v = (2, 2, 2), w = (3, 3, 3) (all collinear, hence coplanar) 5. u = (1, 0, 1), v = (0, 1, 1), w = (1, 1, 2) (since w = u + v) Non-coplanar Vectors: Vectors are non-coplanar if they do not lie in the same plane. For three vectors, they are non-coplanar if their scalar triple product is non-zero (a · (b × c) ≠ 0). Examples: 1. u = i, v = j, w = k 2. u = (1, 0, 0), v = (0, 1, 0), w = (0, 0, 1) 3. u = (1, 1, 0), v = (1, 0, 1), w = (0, 1, 1) 4. u = (1, 2, 3), v = (4, 5, 6), w = (7, 8, 10) 5. u = (1, 0, 0), v = (0, 2, 0), w = (0, 0, 3) 3. Find the magnitude and the unit vector of the following vectors: a) a = 3i - 4j Step 1: Calculate the magnitude of a. |a| = sqrt((3)^2 + (-4)^2 + (0)^2) = sqrt(9 + 16) = sqrt(25) = 5 The magnitude is 5. Step 2: Calculate the unit vector a. a = a|a| = 3i - 4j5 = (3)/(5)i - (4)/(5)j The unit vector is (3)/(5)i - (4)/(5)j. b) b = i + 2j - k Step 1: Calculate the magnitude of b. |b| = sqrt((1)^2 + (2)^2 + (-1)^2) = sqrt(1 + 4 + 1) = sqrt(6) The magnitude is sqrt(6). Step 2: Calculate the unit vector b. b = b|b| = i + 2j - ksqrt(6) = (1)/(sqrt(6))i + (2)/(sqrt(6))j - (1)/(sqrt(6))k The unit vector is (1)/(sqrt(6))i + (2)/(sqrt(6))j - (1)/(sqrt(6))k. c) c = i - 2k (coefficient of j is zero) Step 1: Calculate the magnitude of c. |c| = sqrt((1)^2 + (0)^2 + (-2)^2) = sqrt(1 + 0 + 4) = sqrt(5) The magnitude is sqrt(5). Step 2: Calculate the unit vector c. c = c|c| = i - 2ksqrt(5) = (1)/(sqrt(5))i - (2)/(sqrt(5))k The unit vector is (1)/(sqrt(5))i - (2)/(sqrt(5))k. d) d = 2j - 3k (coefficient of i is zero) Step 1: Calculate the magnitude of d. |d| = sqrt((0)^2 + (2)^2 + (-3)^2) = sqrt(0 + 4 + 9) = sqrt(13) The magnitude is sqrt(13). Step 2: Calculate the unit vector d. d = d|d| = 2j - 3ksqrt(13) = (2)/(sqrt(13))j - (3)/(sqrt(13))k The unit vector is (2)/(sqrt(13))j - (3)/(sqrt(13))k. e) e = 5i - 2j + k Step 1: Calculate the magnitude of e. |e| = sqrt((5)^2 + (-2)^2 + (1)^2) = sqrt(25 + 4 + 1) = sqrt(30) The magnitude is sqrt(30). Step 2: Calculate the unit vector e. e = e|e| = 5i - 2j + ksqrt(30) = (5)/(sqrt(30))i - (2)/(sqrt(30))j + (1)/(sqrt(30))k The unit vector is (5)/(sqrt(30))i - (2)/(sqrt(30))j + (1)/(sqrt(30))k. Got more? Send 'em!