The value of b such that the scalar product of the vector i + j + k with the unit vector parallel to the sum of the vectors 2 i + 4 j - 5 k and b i + 2 j + 3 k is one, is...

Step-by-step solution

Let the sum of the given vectors be v . v = (2 i + 4 j - 5 k ) + (b i + 2 j + 3 k ) = (b+2) i + 6 j - 2 k The unit vector parallel to v is v = (b+2) i + 6 j - 2 k (b+2)^2 + 6^2 + (-2)^2 The scalar product of i + j + k with v is 1 . ( i + j + k ) (b+2) i + 6 j - 2 k (b+2)^2 + 40 = 1 (b+2) + 6 - 2 (b+2)^2 + 40 = 1 b + 6 = (b+2)^2 + 40 Squaring both sides: (b+6)^2 = (b+2)^2 + 40 b^2 + 12b + 36 = b^2 + 4b + 4 + 40 12b + 36 = 4b + 44 8b = 8 b = 1 Answer: 1