Practicing Success
If $\vec{e_1}=\hat i+\hat j+\hat k$ and $\vec{e_2}=\hat i+\hat j-\hat k$ and $\vec a$ and $\vec b$ are two vectors such that $\vec{e_1}=2\vec a+\vec b$ and $\vec{e_2}=\vec a+2\vec b$, then angle between $\vec a$ and $\vec b$ is: |
$\cos^{-1}(\frac{7}{9})$ $\cos^{-1}(\frac{-7}{11})$ $\cos^{-1}(-\frac{7}{9})$ $\cos^{-1}(\frac{6\sqrt{2}}{11})$ |
$\cos^{-1}(\frac{-7}{11})$ |
$2\vec a+\vec b=\hat i+\hat j+\hat k$ $\vec a+2\vec b=\hat i+\hat j-\hat k$ Solve to get: $⇒\vec a=\frac{\hat i}{3}+\frac{\hat j}{3}+\hat k$ and $\vec b=\frac{\hat i}{3}+\frac{\hat j}{3}-\hat k$ $\cos θ=\frac{\vec a.\vec b}{|\vec a||\vec b|}=\frac{(\frac{1}{3})(\frac{1}{3})+(\frac{1}{3})(\frac{1}{3})+(1)(-1)}{(\sqrt{(\frac{1}{3})^2+(\frac{1}{3})^2+1})(\sqrt{(\frac{1}{3})^2+(\frac{1}{3})^2+1})}$ $⇒cos=\frac{-7}{11}⇒θ=cos^{-1}(\frac{-7}{11})$ |