a = i + 5j and b = 5i - j
cosθ\thetaθ = a.b∣a∣.∣b∣\frac{a.b}{|a|.|b|}∣a∣.∣b∣a.b
= (1×5)+(5x−1)(12+52)(52+(−1))2\frac{(1 \times 5)+(5x-1)}{(\sqrt{1^2+5^2})(5^2+(-1))^2}(12+52)(52+(−1))2(1×5)+(5x−1)
= 5−526×26\frac{5-5}{\sqrt{26}\times\sqrt{26}}26×265−5 = 0
x = cos−1^{-1}−1(0), x = 90∘^{\circ}∘
