Question Number 214838 by isaac_newton_2 last updated on 21/Dec/24
$$ \\ $$Determine the unit Vector perpendicular in plane of A = 2i-6j-3k , B = 4i+3j-k
Commented by TonyCWX08 last updated on 21/Dec/24
$${Are}\:{you}\:{sure}\:{A}\:{and}\:{B}\:{are}\:{planes},\:{and}\:{not}\:{vectors}? \\ $$
Commented by mr W last updated on 21/Dec/24
$${he}\:{means}\:{unit}\:{vector}\:{which}\:{is} \\ $$$${perpendicular}\:{to}\:{the}\:{plane}\:{containing} \\ $$$${vectors}\:{A}\:{and}\:{B}. \\ $$
Commented by TonyCWX08 last updated on 21/Dec/24
$${If}\:{that}\:{so},\:{then}\:{alright}. \\ $$
Answered by mr W last updated on 21/Dec/24
$${C}={A}×{B}=\left(\mathrm{2},−\mathrm{6},−\mathrm{3}\right)×\left(\mathrm{4},\mathrm{3},−\mathrm{1}\right) \\ $$$$\:\:\:\:=\left(\mathrm{15},−\mathrm{10},\mathrm{30}\right) \\ $$$$\frac{{C}}{\mid{C}\mid}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{15}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} +\mathrm{30}^{\mathrm{2}} }}\left(\mathrm{15},−\mathrm{10},\mathrm{30}\right) \\ $$$$\:\:\:\:\:\:=\left(\frac{\mathrm{3}}{\mathrm{7}},−\frac{\mathrm{2}}{\mathrm{7}},\frac{\mathrm{6}}{\mathrm{7}}\right)\:\checkmark \\ $$
Commented by mr W last updated on 21/Dec/24
Answered by MrGaster last updated on 21/Dec/24
$$\overset{\rightarrow} {{A}}=\mathrm{2}\hat {{i}}−\mathrm{6}\hat {{j}}−\mathrm{3}\hat {{k}} \\ $$$$\overset{\rightarrow} {{B}}=\mathrm{4}\hat {{i}}+\mathrm{3}\hat {{j}}−\hat {{k}} \\ $$$$\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}=\underset{\mathrm{D}{eterminant}\:{expansi}\mathrm{on}} {\underbrace{\begin{vmatrix}{\hat {{i}}}&{\hat {{j}}}&{\hat {{k}}}\\{\mathrm{2}}&{−\mathrm{6}}&{−\mathrm{3}}\\{\mathrm{4}}&{\mathrm{3}}&{−\mathrm{1}}\end{vmatrix}}}\:=\hat {{i}}\left(\left(−\mathrm{6}\right)\centerdot\left(−\mathrm{1}\right)−\left(−\mathrm{3}\right)\centerdot\mathrm{3}\right)−\hat {{j}}\left(\left(\mathrm{2}\centerdot\left(−\mathrm{1}\right)−\left(−\mathrm{3}\right)\centerdot\mathrm{4}\right)+\hat {{k}}\left(\left(\mathrm{2}\right)\centerdot\mathrm{3}−\left(−\mathrm{6}\right)\centerdot\mathrm{4}\right)\right. \\ $$$$=\hat {{i}}\left(\mathrm{6}+\mathrm{9}\right)−\hat {{j}}\left(−\mathrm{2}+\mathrm{12}\right)+\hat {{k}}\left(\mathrm{6}+\mathrm{24}\right) \\ $$$$=\mathrm{15}\hat {{i}}−\mathrm{10}\hat {{j}}+\mathrm{30}\hat {{k}} \\ $$$$\mid\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\mid=\sqrt{\left(\mathrm{15}\right)^{\mathrm{2}} +\left(\mathrm{10}\right)^{\mathrm{2}} +\left(\mathrm{30}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{225}+\mathrm{100}+\mathrm{900}} \\ $$$$=\sqrt{\mathrm{1225}} \\ $$$$=\mathrm{35} \\ $$$$\hat {{n}}=\frac{\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}}{\mid{A}×{B}\mid} \\ $$$$=\left(\frac{\mathrm{15}}{\mathrm{35}}\right)\hat {{i}}−\left(\frac{\mathrm{10}}{\mathrm{35}}\right)\hat {{j}}+\left(\frac{\mathrm{30}}{\mathrm{35}}\right)\hat {{k}} \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{7}}\right)\hat {{i}}−\left(\frac{\mathrm{2}}{\mathrm{7}}\right)\hat {{j}}+\left(\frac{\mathrm{6}}{\mathrm{7}}\right)\hat {{k}} \\ $$