Question #324969

Find the unit vector that is perpendicular to vectors A=2i + j + k and B= -1 + 2j + k

1
Expert's answer
2022-04-07T14:35:07-0400

n^=A×BA×B\hat{n} =\cfrac{\overrightarrow{A} \times\overrightarrow{B}} {|\overrightarrow{A} \times\overrightarrow{B}|}


A×B=i^j^k^211121=\overrightarrow{A} \times\overrightarrow{B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 1\\ - 1 & 2 & 1 \end{vmatrix}=

=i^(1112)j^(211(1))++k^(221(1))==i^3j^+5k^=\hat{i}(1\cdot1-1\cdot2) -\hat{j}(2\cdot1-1\cdot(-1))+\\ +\hat{k}(2\cdot2-1\cdot(-1))=\\ =-\hat{i}-3\hat{j}+5\hat{k}


A×B=(1)2+(3)2+52=35|\overrightarrow{A} \times\overrightarrow{B}|=\sqrt{(-1)^2+(-3)^2+5^2} =\sqrt{35}


n^=i^353j^35+5k^35.\hat{n} =\cfrac{-\hat{i}} {\sqrt{35}} -\cfrac{3\hat{j}}{\sqrt{35}}+\cfrac{5\hat{k}}{\sqrt{35}}.



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