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Question Number 81295 by jagoll last updated on 11/Feb/20

what is vector unit orthogonal to (1,2,−2) and parallel to yz−plane?

$${what}\:{is}\:{vector}\:{unit}\:{orthogonal} \\ $$$${to}\:\left(\mathrm{1},\mathrm{2},−\mathrm{2}\right)\:{and}\:{parallel}\:{to}\: \\ $$$${yz}−{plane}? \\ $$

Commented by john santu last updated on 11/Feb/20

say this vector b^� = (x,y,z) (i) x^2 +y^2 +z^2 = 1 (ii) b^� . (1,2,−2) = 0 x+2y−2z =0 (iii) b^� parallel to yz−plane ⇒b^� ⊥ i^� ⇒ x =0 then we get y = z = ±(1/2)(√2) ∴ b^� = (0, (1/2)(√2) , (1/2)(√2)) or b^� = (0, −(1/2)(√2) ,−(1/2)(√2) )

$${say}\:{this}\:{vector}\:\bar {{b}}\:=\:\left({x},{y},{z}\right) \\ $$$$\left({i}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\left({ii}\right)\:\bar {{b}}\:.\:\left(\mathrm{1},\mathrm{2},−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$${x}+\mathrm{2}{y}−\mathrm{2}{z}\:=\mathrm{0} \\ $$$$\left({iii}\right)\:\bar {{b}}\:{parallel}\:{to}\:{yz}−{plane} \\ $$$$\Rightarrow\bar {{b}}\:\bot\:\hat {{i}}\:\Rightarrow\:{x}\:=\mathrm{0}\: \\ $$$${then}\:{we}\:{get}\:{y}\:=\:{z}\:=\:\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}} \\ $$$$\therefore\:\bar {{b}}\:=\:\left(\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:,\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\right) \\ $$$${or}\:\bar {{b}}\:=\:\left(\mathrm{0},\:−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:,−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:\right) \\ $$

Commented by jagoll last updated on 11/Feb/20

thx

$$\mathrm{thx} \\ $$

Answered by MJS last updated on 11/Feb/20

parallel to yz−plane = orthogonal to x−axis vector orthogonal to 2 given vectors ((x_1 ),(y_1 ),(z_1 ) ) , ((x_2 ),(y_2 ),(z_2 ) ) can be calculated with a determinant u_x = ((1),(0),(0) ) , u_y = ((0),(1),(0) ) , u_z = ((0),(0),(1) ) determinant ((x_1 ,x_2 ,u_x ),(y_1 ,y_2 ,u_y ),(z_1 ,z_2 ,u_z ))=(y_1 z_2 −y_2 z_1 )u_x +(x_2 z_1 −x_1 z_2 )u_y +(x_1 y_2 −x_2 y_1 )u_z = = (((y_1 z_2 −y_2 z_1 )),((x_2 z_1 −x_1 z_2 )),((x_1 y_2 −x_2 y_1 )) ) in our case this gives ((0),((−2)),((−2)) ) with length 2(√2) ⇒ answer is ± ((0),(((√2)/2)),(((√2)/2)) )

$$\mathrm{parallel}\:\mathrm{to}\:{yz}−\mathrm{plane}\:=\:\mathrm{orthogonal}\:\mathrm{to}\:{x}−\mathrm{axis} \\ $$$$\mathrm{vector}\:\mathrm{orthogonal}\:\mathrm{to}\:\mathrm{2}\:\mathrm{given}\:\mathrm{vectors} \\ $$$$\begin{pmatrix}{{x}_{\mathrm{1}} }\\{{y}_{\mathrm{1}} }\\{{z}_{\mathrm{1}} }\end{pmatrix}\:,\:\begin{pmatrix}{{x}_{\mathrm{2}} }\\{{y}_{\mathrm{2}} }\\{{z}_{\mathrm{2}} }\end{pmatrix}\:\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{determinant} \\ $$$${u}_{{x}} =\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:,\:{u}_{{y}} =\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:,\:{u}_{{z}} =\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{vmatrix}{{x}_{\mathrm{1}} }&{{x}_{\mathrm{2}} }&{{u}_{{x}} }\\{{y}_{\mathrm{1}} }&{{y}_{\mathrm{2}} }&{{u}_{{y}} }\\{{z}_{\mathrm{1}} }&{{z}_{\mathrm{2}} }&{{u}_{{z}} }\end{vmatrix}=\left({y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} \right){u}_{{x}} +\left({x}_{\mathrm{2}} {z}_{\mathrm{1}} −{x}_{\mathrm{1}} {z}_{\mathrm{2}} \right){u}_{{y}} +\left({x}_{\mathrm{1}} {y}_{\mathrm{2}} −{x}_{\mathrm{2}} {y}_{\mathrm{1}} \right){u}_{{z}} = \\ $$$$=\begin{pmatrix}{{y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} {z}_{\mathrm{1}} −{x}_{\mathrm{1}} {z}_{\mathrm{2}} }\\{{x}_{\mathrm{1}} {y}_{\mathrm{2}} −{x}_{\mathrm{2}} {y}_{\mathrm{1}} }\end{pmatrix} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:\mathrm{this}\:\mathrm{gives} \\ $$$$\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{2}}\\{−\mathrm{2}}\end{pmatrix}\:\mathrm{with}\:\mathrm{length}\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\pm\begin{pmatrix}{\mathrm{0}}\\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{pmatrix} \\ $$

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