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Question Number 132972 by rexford last updated on 17/Feb/21

Answered by Ar Brandon last updated on 17/Feb/21

$$\left|\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& -2& 3\\ 1& -1& -2\end{array}\right|=7\mathrm{i}+5\mathrm{j}+\mathrm{k}$$

Commented by rexford last updated on 17/Feb/21

$$please,isthereanyexplanationbehind$$$$theuseofthedeterminant$$

Commented by mr W last updated on 17/Feb/21

$$\mathit{a}\times \mathit{b}isperpendicularbothto\mathit{a}and$$$$to\mathit{b}.$$$$\mathit{a}\times \mathit{b}=\left|\begin{array}{ccc}i& j& k\\ a_i& a_j& a_k\\ b_i& b_j& b_k\end{array}\right|$$

Answered by liberty last updated on 17/Feb/21

$$\mathrm{your}\mathrm{asked}\mathrm{unit}\mathrm{vector}$$$$\mathrm{call}{\mathrm{it}}^{\prime }\mathrm{s}\overrightarrow{c}\mathrm{where}\overrightarrow{c}=\frac{\left|\begin{array}{c}1-23\\ 1-1-2\end{array}\right|}{\mid \mid \overrightarrow{c}\mid \mid }$$$$\overrightarrow{c}=\frac{7\hat{\mathrm{i}}+5\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{49+25+1}}=\frac{7\hat{\mathrm{i}}+5\hat{\mathrm{j}}+\hat{\mathrm{k}}}{5\sqrt{3}}$$$$$$

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