(a) Is the vector a unit vector? Justify your answer. (b) Can a unit vector have any components with magnitude greater than unity? Can it have any negative components? In each case justify your answer. (c) If where a? ? is a constant, determine the value of a that makes a unit vector.
Solution 44E Problem (a) Step 1: Consider a =i + j + k ˆ The given vector is not a unit vector, because modulus of unit vector must be equal to 1. Proof 2 2 2 | | 1 + 1 + 1 | | 1 + 1 + 1 | | 3 The given has the modulus greater than unity, hence it is not a unit vector Problem (b) Step 1: The unit vector can not have the components greater than 1. Justification Sum of squares of vector components will be having magnitude greater than 1. Even if we take square root of the value found, we can not get value equal to 1. EX. consider vector 2i +3 j + 4kˆ 2 2 2 Modulus of the vector taken = 2 + 3 + 4 = 4 + 9 + 16 modulus = 29 which is greater than 1. Hence unit vector can not have magnitudes greater than 1. Step 2: The unit vector can have negative components. Justification 3 ˆ 4 ˆ Take a vector a = i 5 5 j 2 2 Modulus of the above vector = ( 5 + ( )45 9 16 = 25+ 25 25 = 25 modulus = 1 Since the modulus of the vector taken is 1, we can say that the unit vectors can have negative components. Problem ( c ) Step 1: Given vector A = a (3.0i + 4.0j ) Where a is constant In order to make vector as unit vector, modulus of A must be equal to 1. Step 2: ˆ ˆ A = a (3.0i + 4.0j ) A =3.0ai + 4.0ajˆ (3a) + (4a)2 Therefore modulus of = 2 2 = 9a + 16a 2 Taking a outside the square root = a 9 + 16 = a 25 Modulus of A = 5a2