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Question Number 86406 by mathmax by abdo last updated on 28/Mar/20

let u^→ =i^→ −j^→ +k^→ and v^→ =2i^→ +j^→ +3k^→ (o,i,j,k) orthonormal 1) calculate ∣∣u^→ ∣∣ ,∣∣v^→ ∣∣ ,u^→ .v^→ 2) calculate cos(u^→ ,v^→ ) 3)calculate u^→ Λv^→ and sin(u^→ ,v^→ )

$${let}\:\:\overset{\rightarrow} {{u}}=\overset{\rightarrow} {{i}}−\overset{\rightarrow} {{j}}\:+\overset{\rightarrow} {{k}}\:{and}\:\overset{\rightarrow} {{v}}=\mathrm{2}\overset{\rightarrow} {{i}}+\overset{\rightarrow} {{j}}\:+\mathrm{3}\overset{\rightarrow} {{k}} \\ $$$$\left({o},{i},{j},{k}\right)\:{orthonormal} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:\mid\mid\overset{\rightarrow} {{u}}\mid\mid\:\:,\mid\mid\overset{\rightarrow} {{v}}\mid\mid\:\:,\overset{\rightarrow} {{u}}.\overset{\rightarrow} {{v}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{cos}\left(\overset{\rightarrow} {{u}},\overset{\rightarrow} {{v}}\right) \\ $$$$\left.\mathrm{3}\right){calculate}\:\overset{\rightarrow} {{u}}\Lambda\overset{\rightarrow} {{v}}\:\:\:\:{and}\:\:{sin}\left(\overset{\rightarrow} {{u}},\overset{\rightarrow} {{v}}\right) \\ $$

Answered by jagoll last updated on 28/Mar/20

∣∣u^→ ∣∣ = (√3) ∣∣v^→ ∣∣ = (√(14)) u^→ . v^→ = 2−1+3 = 4 cos (u^→ , v^→ ) = (4/(√(42)))

$$\mid\mid\overset{\rightarrow} {\mathrm{u}}\mid\mid\:=\:\sqrt{\mathrm{3}} \\ $$$$\mid\mid\overset{\rightarrow} {\mathrm{v}}\mid\mid\:=\:\sqrt{\mathrm{14}} \\ $$$$\overset{\rightarrow} {\mathrm{u}}\:.\:\overset{\rightarrow} {\mathrm{v}}\:=\:\mathrm{2}−\mathrm{1}+\mathrm{3}\:=\:\mathrm{4} \\ $$$$\mathrm{cos}\:\left(\overset{\rightarrow} {\mathrm{u}},\:\overset{\rightarrow} {\mathrm{v}}\right)\:=\:\frac{\mathrm{4}}{\sqrt{\mathrm{42}}} \\ $$

Answered by Rio Michael last updated on 28/Mar/20

3)∣ u^→ × v^→ ∣ = ∣u^→ ∣ ∣v^→ ∣ sinθ ⇒ sin θ = ((∣u^→ ×v^→ ∣)/(∣u^→ ∣∣v^→ ∣)) u^→ ×v^→ = determinant ((i^→ ,j^→ ,k^→ ),(1,(−1),1),(2,1,3)) = i determinant (((−1),1),(1,3))−j determinant ((1,1),(2,3))+ k determinant ((1,(−1)),(2,1)) = −4i −j + 3k ∣u^→ ×v^→ ∣ = (√(26)) ∣u^→ ∣ = (√3) and ∣v^→ ∣ = (√(14)) ⇒ sin (u^→ ,v^→ ) = ((√(26))/((√3) (√(14)))) = (√((13)/(14)))

$$\left.\:\mathrm{3}\right)\mid\:\overset{\rightarrow} {{u}}\:×\:\overset{\rightarrow} {{v}}\mid\:=\:\mid\overset{\rightarrow} {{u}}\mid\:\mid\overset{\rightarrow} {{v}}\mid\:\mathrm{sin}\theta \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{sin}\:\theta\:=\:\frac{\mid\overset{\rightarrow} {{u}}\:×\overset{\rightarrow} {{v}}\mid}{\mid\overset{\rightarrow} {{u}}\mid\mid\overset{\rightarrow} {{v}}\mid} \\ $$$$\:\:\:\:\:\overset{\rightarrow} {{u}}×\overset{\rightarrow} {{v}}\:=\:\begin{vmatrix}{\overset{\rightarrow} {{i}}}&{\overset{\rightarrow} {{j}}}&{\overset{\rightarrow} {{k}}}\\{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\end{vmatrix}\:=\:{i}\begin{vmatrix}{−\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{3}}\end{vmatrix}−{j}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{3}}\end{vmatrix}+\:{k}\begin{vmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{\mathrm{2}}&{\mathrm{1}}\end{vmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\mathrm{4}{i}\:−{j}\:+\:\mathrm{3}{k} \\ $$$$\:\:\:\:\:\mid\overset{\rightarrow} {{u}}×\overset{\rightarrow} {{v}}\mid\:=\:\sqrt{\mathrm{26}} \\ $$$$\:\:\mid\overset{\rightarrow} {{u}}\mid\:=\:\sqrt{\mathrm{3}}\:\:\mathrm{and}\:\mid\overset{\rightarrow} {{v}}\mid\:=\:\sqrt{\mathrm{14}}\:\:\:\Rightarrow\:\mathrm{sin}\:\left(\overset{\rightarrow} {{u}},\overset{\rightarrow} {{v}}\right)\:=\:\frac{\sqrt{\mathrm{26}}}{\sqrt{\mathrm{3}}\:\sqrt{\mathrm{14}}}\:=\:\sqrt{\frac{\mathrm{13}}{\mathrm{14}}} \\ $$

Commented by abdomathmax last updated on 28/Mar/20

thankx sir.

$${thankx}\:{sir}. \\ $$

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