Question Number 78889 by gopikrishnan last updated on 21/Jan/20
$${the}\:{angle}\:{between}\:{the}\:{planes}\:{r}^{\rightarrow} .\left(\mathrm{2}{i}^{\rightarrow} +\mathrm{2}{j}^{\rightarrow} +\mathrm{2}{k}^{\rightarrow} \right)=\mathrm{4}\:{and}\:\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{2}{z}=\mathrm{15}\:{is} \\ $$
Answered by Kunal12588 last updated on 21/Jan/20
$$\pi_{\mathrm{1}} \::\:\overset{\rightarrow} {{r}}\centerdot\left(\mathrm{2}{i}+\mathrm{2}{j}+\mathrm{2}{k}\right)=\mathrm{4} \\ $$$$\pi_{\mathrm{2}} :\:\overset{\rightarrow} {{r}}\centerdot\left(\mathrm{4}{i}−\mathrm{2}{j}+\mathrm{2}{k}\right)=\mathrm{15} \\ $$$${you}\:{just}\:{have}\:{to}\:{find}\:{the}\:{angle}\:{between}\:{the} \\ $$$${vectors}\:\left(\mathrm{2}{i}+\mathrm{2}{j}+\mathrm{2}{k}\right)\:{and}\:\left(\mathrm{4}{i}−\mathrm{2}{j}+\mathrm{2}{k}\right). \\ $$