Question Number 91496 by Zainal Arifin last updated on 01/May/20
$$\mathrm{The}\:\mathrm{vector}\:\boldsymbol{\mathrm{a}}=\mathrm{3}\boldsymbol{\mathrm{i}}−\mathrm{2}\boldsymbol{\mathrm{j}}+\mathrm{2}\boldsymbol{\mathrm{k}}\:\:\mathrm{and}\:\boldsymbol{\mathrm{b}}=−\boldsymbol{\mathrm{i}}−\mathrm{2}\boldsymbol{\mathrm{k}} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{adjacent}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{parallelogram}. \\ $$$$\mathrm{Then}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{its}\:\mathrm{diagonal}\:\mathrm{is} \\ $$
Commented by jagoll last updated on 01/May/20
$$\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\:=\:\left(\mathrm{2},−\mathrm{2},\mathrm{0}\right) \\ $$$$\overset{\rightarrow} {{a}}−\overset{\rightarrow} {{b}}\:=\:\left(\mathrm{4},−\mathrm{2},\:\mathrm{4}\right) \\ $$$${let}\:\theta\:=\:{the}\:{angle}\:{between}\:{diagonals} \\ $$$$\mathrm{cos}\:\theta\:=\:\frac{\mathrm{8}+\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{2}}\:.\mathrm{6}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\Rightarrow\theta=\frac{\pi}{\mathrm{4}} \\ $$