Question

Find the magnitude of the dot product of the two vectors B=6i+j+2k,D=-i+3j.what is the angle between B and D.

Answer 1

$B=6i+j+2k \newline D=-i+3j \newline B.D=(6i+j+2k ).(-i+3j) \newline B.D=(-6+3)=-3 \\~\\$

From the equation,

$B.D=|B||D|cos(\theta)$ where $\theta$ is the angle between two vectors

$\cos(\theta)=\frac{-3}{(\sqrt{41})(\sqrt{10})} \newline \theta=cos^{-1}(\frac{-3}{\sqrt{410}}) \newline \theta=98.52\degree$