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Question Number 182176 by peter frank last updated on 05/Dec/22

Answered by MikeH last updated on 05/Dec/22

let w^→ = xi + 6j + y k orthogonal to u^→ and v^→ ⇒ w^→ = u^→ × v^→ ⇒ xi + 6j + yk = determinant ((i,j,k),((−1),0,1),((−1),(−2),2)) ⇒ xi + 6j + yk = i(2)−j(−1) + k(2) ⇒ xi + 6j + yk = 2i + j + 2k x = 2 and y = 2 ⇒ w^→ = 2i + 6j + 2k

$$\mathrm{let}\:\overset{\rightarrow} {\mathrm{w}}\:=\:{x}\mathrm{i}\:+\:\mathrm{6j}\:+\:{y}\:\mathrm{k} \\ $$$$\mathrm{orthogonal}\:\mathrm{to}\:\overset{\rightarrow} {\mathrm{u}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{v}}\:\Rightarrow\:\overset{\rightarrow} {\mathrm{w}}\:=\:\overset{\rightarrow} {\mathrm{u}}×\:\overset{\rightarrow} {\mathrm{v}} \\ $$$$\Rightarrow\:{x}\mathrm{i}\:+\:\mathrm{6j}\:+\:{y}\mathrm{k}\:=\:\begin{vmatrix}{\mathrm{i}}&{\mathrm{j}}&{\mathrm{k}}\\{−\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\\{−\mathrm{1}}&{−\mathrm{2}}&{\mathrm{2}}\end{vmatrix} \\ $$$$\Rightarrow\:{x}\mathrm{i}\:+\:\mathrm{6j}\:+\:{y}\mathrm{k}\:=\:\mathrm{i}\left(\mathrm{2}\right)−\mathrm{j}\left(−\mathrm{1}\right)\:+\:\mathrm{k}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:{x}\mathrm{i}\:+\:\mathrm{6j}\:+\:{y}\mathrm{k}\:=\:\mathrm{2i}\:+\:\mathrm{j}\:+\:\mathrm{2k} \\ $$$${x}\:=\:\mathrm{2}\:\mathrm{and}\:{y}\:=\:\mathrm{2} \\ $$$$\Rightarrow\:\overset{\rightarrow} {\mathrm{w}}\:=\:\mathrm{2i}\:+\:\mathrm{6j}\:+\:\mathrm{2k}\: \\ $$

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