In OAB , O(0,0,0), A(6,2,-3) and B(4,0,3) are the vertices. Let a and b be position vectors of points A and B respectively and OM is the projection of a on b then l(AM) is equal to...

Step-by-step solution

Given position vectors a = 6 i + 2 j - 3 k and b = 4 i + 3 k . The magnitude of a is: | a | = 6^2 + 2^2 + (-3)^2 = 36 + 4 + 9 = 49 = 7 The magnitude of b is: | b | = 4^2 + 0^2 + 3^2 = 16 + 0 + 9 = 25 = 5 The length of the projection of a on b is OM , which is given by: OM = a b | b | Calculating the dot product: a b = (6)(4) + (2)(0) + (-3)(3) = 24 + 0 - 9 = 15 Substituting the values to find OM : OM = 15 5 = 3 In the right-angled OAM , OA is the hypotenuse with length | a | . Using the Pythagoras theorem: OA^2 = O