Question Number 81295 by jagoll last updated on 11/Feb/20
$${what}\:{is}\:{vector}\:{unit}\:{orthogonal} \\ $$$${to}\:\left(\mathrm{1},\mathrm{2},−\mathrm{2}\right)\:{and}\:{parallel}\:{to}\: \\ $$$${yz}−{plane}? \\ $$
Commented by john santu last updated on 11/Feb/20
$${say}\:{this}\:{vector}\:\bar {{b}}\:=\:\left({x},{y},{z}\right) \\ $$$$\left({i}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\left({ii}\right)\:\bar {{b}}\:.\:\left(\mathrm{1},\mathrm{2},−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$${x}+\mathrm{2}{y}−\mathrm{2}{z}\:=\mathrm{0} \\ $$$$\left({iii}\right)\:\bar {{b}}\:{parallel}\:{to}\:{yz}−{plane} \\ $$$$\Rightarrow\bar {{b}}\:\bot\:\hat {{i}}\:\Rightarrow\:{x}\:=\mathrm{0}\: \\ $$$${then}\:{we}\:{get}\:{y}\:=\:{z}\:=\:\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}} \\ $$$$\therefore\:\bar {{b}}\:=\:\left(\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:,\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\right) \\ $$$${or}\:\bar {{b}}\:=\:\left(\mathrm{0},\:−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:,−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:\right) \\ $$
Commented by jagoll last updated on 11/Feb/20
$$\mathrm{thx} \\ $$
Answered by MJS last updated on 11/Feb/20
$$\mathrm{parallel}\:\mathrm{to}\:{yz}−\mathrm{plane}\:=\:\mathrm{orthogonal}\:\mathrm{to}\:{x}−\mathrm{axis} \\ $$$$\mathrm{vector}\:\mathrm{orthogonal}\:\mathrm{to}\:\mathrm{2}\:\mathrm{given}\:\mathrm{vectors} \\ $$$$\begin{pmatrix}{{x}_{\mathrm{1}} }\\{{y}_{\mathrm{1}} }\\{{z}_{\mathrm{1}} }\end{pmatrix}\:,\:\begin{pmatrix}{{x}_{\mathrm{2}} }\\{{y}_{\mathrm{2}} }\\{{z}_{\mathrm{2}} }\end{pmatrix}\:\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{determinant} \\ $$$${u}_{{x}} =\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:,\:{u}_{{y}} =\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:,\:{u}_{{z}} =\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{vmatrix}{{x}_{\mathrm{1}} }&{{x}_{\mathrm{2}} }&{{u}_{{x}} }\\{{y}_{\mathrm{1}} }&{{y}_{\mathrm{2}} }&{{u}_{{y}} }\\{{z}_{\mathrm{1}} }&{{z}_{\mathrm{2}} }&{{u}_{{z}} }\end{vmatrix}=\left({y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} \right){u}_{{x}} +\left({x}_{\mathrm{2}} {z}_{\mathrm{1}} −{x}_{\mathrm{1}} {z}_{\mathrm{2}} \right){u}_{{y}} +\left({x}_{\mathrm{1}} {y}_{\mathrm{2}} −{x}_{\mathrm{2}} {y}_{\mathrm{1}} \right){u}_{{z}} = \\ $$$$=\begin{pmatrix}{{y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} {z}_{\mathrm{1}} −{x}_{\mathrm{1}} {z}_{\mathrm{2}} }\\{{x}_{\mathrm{1}} {y}_{\mathrm{2}} −{x}_{\mathrm{2}} {y}_{\mathrm{1}} }\end{pmatrix} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:\mathrm{this}\:\mathrm{gives} \\ $$$$\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{2}}\\{−\mathrm{2}}\end{pmatrix}\:\mathrm{with}\:\mathrm{length}\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\pm\begin{pmatrix}{\mathrm{0}}\\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{pmatrix} \\ $$