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Question Number 76524 by john santu last updated on 28/Dec/19
find vector unit perpendicular to vector a^− =(1,2,3) and b^− =(−1,0,2)
$${find}\:{vector}\:{unit}\:{perpendicular}\: \\ $$$${to}\:{vector}\:\overset{−} {{a}}=\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{and}\:\overset{−} {{b}}=\left(−\mathrm{1},\mathrm{0},\mathrm{2}\right) \\ $$
Answered by MJS last updated on 28/Dec/19
determinant ((1,(−1),u_x ),(2,0,u_y ),(3,2,u_z ))=4u_x −5u_y +2u_z = ((4),((−5)),(2) ) abs ((4),((−5)),(2) ) =3(√5) (1/(3(√5))) ((4),((−5)),(2) ) = ((((4(√5))/(15))),((−((√5)/3))),(((2(√5))/(15))) )
$$\begin{vmatrix}{\mathrm{1}}&{−\mathrm{1}}&{{u}_{{x}} }\\{\mathrm{2}}&{\mathrm{0}}&{{u}_{{y}} }\\{\mathrm{3}}&{\mathrm{2}}&{{u}_{{z}} }\end{vmatrix}=\mathrm{4}{u}_{{x}} −\mathrm{5}{u}_{{y}} +\mathrm{2}{u}_{{z}} =\begin{pmatrix}{\mathrm{4}}\\{−\mathrm{5}}\\{\mathrm{2}}\end{pmatrix} \\ $$$$\mathrm{abs}\:\begin{pmatrix}{\mathrm{4}}\\{−\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}\:=\mathrm{3}\sqrt{\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{5}}}\begin{pmatrix}{\mathrm{4}}\\{−\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}\:=\begin{pmatrix}{\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{15}}}\\{−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}}\\{\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{15}}}\end{pmatrix} \\ $$
Commented by benjo 1/2 santuyy last updated on 28/Dec/19
why sir not make ±?
$${why}\:{sir}\:{not}\:{make}\:\pm? \\ $$
Commented by MJS last updated on 28/Dec/19
of course both + and − are ok, I thought one of these is enough...
$$\mathrm{of}\:\mathrm{course}\:\mathrm{both}\:+\:\mathrm{and}\:−\:\mathrm{are}\:\mathrm{ok},\:\mathrm{I}\:\mathrm{thought}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{is}\:\mathrm{enough}… \\ $$
Commented by benjo 1/2 santuyy last updated on 28/Dec/19
oo ok sir thanks you
$${oo}\:{ok}\:{sir}\:{thanks}\:{you} \\ $$
Answered by vishalbhardwaj last updated on 28/Dec/19
a^→ =i^� +2j^� +3k^� and b^→ =−i^� +2k^� a^→ ×b^→ =(i^� +2j^� +3k^� )×(−i^� +2k^� ) = −2(i^� ×k^� )−2(j^� ×i^� )+4(j^� ×k^� ) −3(k^� ×i^� ) = 2j^� +2k^� +4i^� −3j^� = 4i^� −j^� +2k^� and ∣a^→ ×b^→ ∣=(√((4)^2 +(−1)^2 +(2)^2 )) = (√(16+1+4))= (√(21)) let p^→ =a^→ ×b^→ , p^� = (p^→ /(∣p^→ ∣)) = ((4i^� −j^� +2k^� )/( (√(21)))) ⇒ p^� = (4/( (√(21)))) i^� −(1/( (√(21)))) j^� +(2/( (√(21)))) k^�
$$\overset{\rightarrow} {{a}}=\hat {{i}}+\mathrm{2}\hat {{j}}+\mathrm{3}\hat {{k}}\:\:\mathrm{and}\:\overset{\rightarrow} {{b}}=−\hat {{i}}+\mathrm{2}\hat {{k}} \\ $$$$\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}=\left(\hat {{i}}+\mathrm{2}\hat {{j}}+\mathrm{3}\hat {{k}}\right)×\left(−\hat {{i}}+\mathrm{2}\hat {{k}}\right) \\ $$$$=\:−\mathrm{2}\left(\hat {{i}}×\hat {{k}}\right)−\mathrm{2}\left(\hat {{j}}×\hat {{i}}\right)+\mathrm{4}\left(\hat {{j}}×\hat {{k}}\right) \\ $$$$−\mathrm{3}\left(\hat {{k}}×\hat {{i}}\right) \\ $$$$=\:\mathrm{2}\hat {{j}}+\mathrm{2}\hat {{k}}+\mathrm{4}\hat {{i}}−\mathrm{3}\hat {{j}} \\ $$$$=\:\mathrm{4}\hat {{i}}−\hat {{j}}+\mathrm{2}\hat {{k}} \\ $$$$\mathrm{and}\:\:\:\mid\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\mid=\sqrt{\left(\mathrm{4}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\:\sqrt{\mathrm{16}+\mathrm{1}+\mathrm{4}}=\:\sqrt{\mathrm{21}} \\ $$$$\mathrm{let}\:\overset{\rightarrow} {{p}}=\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}},\:\:\:\hat {{p}}\:=\:\frac{\overset{\rightarrow} {{p}}}{\mid\overset{\rightarrow} {{p}}\mid}\:=\:\frac{\mathrm{4}\hat {{i}}−\hat {{j}}+\mathrm{2}\hat {{k}}}{\:\sqrt{\mathrm{21}}} \\ $$$$\Rightarrow\:\:\:\hat {{p}}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{21}}}\:\hat {{i}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{21}}}\:\hat {{j}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{21}}}\:\hat {{k}} \\ $$
Commented by benjo 1/2 santuyy last updated on 28/Dec/19
sorry sir. but wrong a×b
$${sorry}\:{sir}.\:{but}\:{wrong}\:{a}×{b}\: \\ $$
Commented by benjo 1/2 santuyy last updated on 28/Dec/19
i got a × b = 4i − 5j +2k
$${i}\:{got}\:{a}\:×\:{b}\:=\:\mathrm{4}{i}\:−\:\mathrm{5}{j}\:+\mathrm{2}{k}\: \\ $$
Commented by JDamian last updated on 28/Dec/19
i×2k ≠ −2(i×k) i×2k = −2j santuyy is right
$$ \\ $$$${i}×\mathrm{2}{k}\:\:\neq\:\:−\mathrm{2}\left({i}×{k}\right) \\ $$$${i}×\mathrm{2}{k}\:\:=\:\:−\mathrm{2}{j} \\ $$$$ \\ $$$${santuyy}\:\:\:{is}\:\:\:{right} \\ $$
Commented by benjo 1/2 santuyy last updated on 28/Dec/19
thanks you sir
$${thanks}\:{you}\:{sir} \\ $$
Answered by vishalbhardwaj last updated on 28/Dec/19
a^→ =i^� +2j^� +3k^� and b^→ =−i^� +2k^� a^→ ×b^→ =(i^� +2j^� +3k^� )×(−i^� +2k^� ) = −2(i^� ×k^� )−2(j^� ×i^� )+4(j^� ×k^� ) −3(k^� ×i^� ) = 2j^� +2k^� +4i^� −3j^� = 4i^� −j^� +2k^� and ∣a^→ ×b^→ ∣=(√((4)^2 +(−1)^2 +(2)^2 )) = (√(16+1+4))= (√(21)) let p^→ =a^→ ×b^→ , p^� = (p^→ /(∣p^→ ∣)) = ((4i^� −j^� +2k^� )/( (√(21)))) ⇒ p^� = (4/( (√(21)))) i^� −(1/( (√(21)))) j^� +(2/( (√(21)))) k^�
$$\overset{\rightarrow} {{a}}=\hat {{i}}+\mathrm{2}\hat {{j}}+\mathrm{3}\hat {{k}}\:\:\mathrm{and}\:\overset{\rightarrow} {{b}}=−\hat {{i}}+\mathrm{2}\hat {{k}} \\ $$$$\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}=\left(\hat {{i}}+\mathrm{2}\hat {{j}}+\mathrm{3}\hat {{k}}\right)×\left(−\hat {{i}}+\mathrm{2}\hat {{k}}\right) \\ $$$$=\:−\mathrm{2}\left(\hat {{i}}×\hat {{k}}\right)−\mathrm{2}\left(\hat {{j}}×\hat {{i}}\right)+\mathrm{4}\left(\hat {{j}}×\hat {{k}}\right) \\ $$$$−\mathrm{3}\left(\hat {{k}}×\hat {{i}}\right) \\ $$$$=\:\mathrm{2}\hat {{j}}+\mathrm{2}\hat {{k}}+\mathrm{4}\hat {{i}}−\mathrm{3}\hat {{j}} \\ $$$$=\:\mathrm{4}\hat {{i}}−\hat {{j}}+\mathrm{2}\hat {{k}} \\ $$$$\mathrm{and}\:\:\:\mid\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\mid=\sqrt{\left(\mathrm{4}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\:\sqrt{\mathrm{16}+\mathrm{1}+\mathrm{4}}=\:\sqrt{\mathrm{21}} \\ $$$$\mathrm{let}\:\overset{\rightarrow} {{p}}=\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}},\:\:\:\hat {{p}}\:=\:\frac{\overset{\rightarrow} {{p}}}{\mid\overset{\rightarrow} {{p}}\mid}\:=\:\frac{\mathrm{4}\hat {{i}}−\hat {{j}}+\mathrm{2}\hat {{k}}}{\:\sqrt{\mathrm{21}}} \\ $$$$\Rightarrow\:\:\:\hat {{p}}\:=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{21}}}\:\hat {{i}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{21}}}\:\hat {{j}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{21}}}\:\hat {{k}} \\ $$

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