Find the unit vector that is perpendicular to vectors A=2i + j + k and B= -1 + 2j + k
n^=A→×B→∣A→×B→∣\hat{n} =\cfrac{\overrightarrow{A} \times\overrightarrow{B}} {|\overrightarrow{A} \times\overrightarrow{B}|}n^=∣A×B∣A×B
A→×B→=∣i^j^k^211−121∣=\overrightarrow{A} \times\overrightarrow{B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & 1\\ - 1 & 2 & 1 \end{vmatrix}=A×B=∣∣i^2−1j^12k^11∣∣=
=i^(1⋅1−1⋅2)−j^(2⋅1−1⋅(−1))++k^(2⋅2−1⋅(−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}=i^(1⋅1−1⋅2)−j^(2⋅1−1⋅(−1))++k^(2⋅2−1⋅(−1))==−i^−3j^+5k^
∣A→×B→∣=(−1)2+(−3)2+52=35|\overrightarrow{A} \times\overrightarrow{B}|=\sqrt{(-1)^2+(-3)^2+5^2} =\sqrt{35}∣A×B∣=(−1)2+(−3)2+52=35
n^=−i^35−3j^35+5k^35.\hat{n} =\cfrac{-\hat{i}} {\sqrt{35}} -\cfrac{3\hat{j}}{\sqrt{35}}+\cfrac{5\hat{k}}{\sqrt{35}}.n^=35−i^−353j^+355k^.
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