Find the value of ' b ' such that the scalar product of the vector + + k with the unit vector parallel to the sum of the vectors 2 +4 -5 k and b +2 +3 k is unity

Step-by-step solution

Let a = i + j + k . Let u = 2 i + 4 j - 5 k and v = b i + 2 j + 3 k . The sum of the vectors is s = u + v = (2+b) i + 6 j - 2 k . The unit vector parallel to s is s = s | s | = (2+b) i + 6 j - 2 k (2+b)^2 + 6^2 + (-2)^2 . The scalar product of a and s is given as unity: a s = 1 (1)(2+b) + (1)(6) + (1)(-2) (2+b)^2 + 36 + 4 = 1 . Simplifying the numerator: 2 + b + 6 - 2 = b + 6 . So, b + 6 = (2+b)^2 + 40 . Squaring both sides: (b+6)^2 = (2+b)^2 + 40 . b^2 + 12b + 36 = b^2 + 4b + 4 + 40 . 12b + 36 = 4b + 44 . 8b = 8 .