Question Number 170386 by Mastermind last updated on 22/May/22
$${Find}\:{the}\:{volume}\:{of}\:{tetrahedron}\:{whose} \\ $$$${vertices}\:{are}\:{the}\:{points}\:{A}\left(\mathrm{2},\:−\mathrm{1},\:−\mathrm{3}\right), \\ $$$${B}\left(\mathrm{4},\:\mathrm{1},\:\mathrm{3}\right),\:{C}\left(\mathrm{3},\:\mathrm{2},\:−\mathrm{1}\right)\:{and}\:{D}\left(\mathrm{1},\:\mathrm{4},\:\mathrm{2}\right). \\ $$$$ \\ $$$${Mastermind} \\ $$
Commented by mr W last updated on 22/May/22
$$\boldsymbol{{a}}=\overset{\rightarrow} {{AD}}=\left(−\mathrm{1},\mathrm{5},\mathrm{4}\right) \\ $$$$\boldsymbol{{b}}=\overset{\rightarrow} {{BD}}=\left(−\mathrm{3},\mathrm{3},−\mathrm{1}\right) \\ $$$$\boldsymbol{{c}}=\overset{\rightarrow} {{CD}}=\left(−\mathrm{2},\mathrm{2},\mathrm{3}\right) \\ $$$${V}=\frac{\mid\boldsymbol{{a}}×\boldsymbol{{b}}\centerdot\boldsymbol{{c}}\mid}{\mathrm{6}} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{6}}\begin{vmatrix}{−\mathrm{1}}&{\mathrm{5}}&{\mathrm{4}}\\{−\mathrm{3}}&{\mathrm{3}}&{−\mathrm{1}}\\{−\mathrm{2}}&{\mathrm{2}}&{\mathrm{3}}\end{vmatrix}=\frac{\mathrm{44}}{\mathrm{6}}=\frac{\mathrm{22}}{\mathrm{3}} \\ $$
Commented by Mastermind last updated on 22/May/22
$${got}\:{same} \\ $$$${thank}\:{you} \\ $$
Answered by MikeH last updated on 23/May/22
$${V}\:=\frac{\mathrm{1}}{\mathrm{6}}\begin{vmatrix}{\mathrm{2}}&{−\mathrm{1}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{1}}&{\mathrm{3}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{4}}&{\mathrm{2}}&{\mathrm{1}}\end{vmatrix}=\frac{\mathrm{22}}{\mathrm{3}} \\ $$$$ \\ $$