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Question Number 120801 by TITA last updated on 02/Nov/20

Commented by liberty last updated on 02/Nov/20

$$\mathrm{let}\{\begin{array}{l}\mathrm{i}=(1,0,0)\\ \mathrm{j}=(0,1,0)\\ \mathrm{k}=(0,0,1)\end{array}\to \{\begin{array}{l}\mathrm{i}.\mathrm{i}=1.1+0.0+0.0=1\\ \mathrm{j}.\mathrm{j}=0.0+1.1+0.0=1\\ \mathrm{k}.\mathrm{k}=0.0+0.0+1.1=1\end{array}$$

Commented by TITA last updated on 03/Nov/20

$$thanks$$

Answered by $@y@m last updated on 03/Nov/20

$$\overrightarrow{a}.\overrightarrow{b}=\mid a\mid \mid b\mid \cos \theta$$$$\left(i\right)Put\overrightarrow{a}=\overrightarrow{b}=\hat{i}$$$$\Rightarrow \hat{i}.\hat{i}=\mid \hat{i}\mid \mid \hat{i}\mid \cos 0=1andsoon.$$$$\left(ii\right)Put\overrightarrow{a}=\hat{i},\overrightarrow{b}=\hat{j}$$$$\Rightarrow \hat{i}.\hat{j}=\mid \hat{i}\mid \mid \hat{j}\mid \cos 90=0andsoon.$$

Commented by TITA last updated on 03/Nov/20

$$thanks$$

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