Question

find the angle made by the vector 4i+3j+5k on x axis​

Answer 1

Given a vector ${\bf A}=4\hat i+3\hat j+5\hat k.$ The angle between ${\bf A}$ and positive $x$ axis is given by

$$\theta=\cos^{-1}\frac{A_x}{A}=\cos^{-1}\frac{A_x}{\sqrt{A_x^2+A_y^2+A_z^2}}$$ $$\theta=\cos^{-1}\frac{4}{\sqrt{4^2+3^2+5^2}}=\cos^{-1}\frac{4}{5\sqrt{2}}=0.566=55.6^{\circ}$$