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Question Number 147162 by nadovic last updated on 18/Jul/21

Answered by liberty last updated on 18/Jul/21

$$\left(1\right)\overrightarrow{p}=\frac{2}{5}\overrightarrow{a}+\frac{1}{5}\overrightarrow{b}+\frac{2}{5}\overrightarrow{c}$$$$\left(2\right)\overrightarrow{p}.\overrightarrow{b}=\overrightarrow{p}.\overrightarrow{a}$$$$\frac{2}{5}\overrightarrow{a}.\overrightarrow{b}+\frac{1}{5}\mid \overrightarrow{b}{\mid }^2+\frac{2}{5}\overrightarrow{b}.\overrightarrow{c}=$$$$\frac{2}{5}\mid \overrightarrow{a}{\mid }^2+\frac{1}{5}\overrightarrow{a}.\overrightarrow{b}+\frac{2}{5}\overrightarrow{a}.\overrightarrow{c}$$$$\frac{2}{5}\overrightarrow{b}.\overrightarrow{c}+\frac{1}{5}\mid \overrightarrow{b}{\mid }^2=\frac{2}{5}\mid \overrightarrow{a}{\mid }^2+\frac{2}{5}\overrightarrow{a}.\overrightarrow{c}$$$$2\overrightarrow{b}.\overrightarrow{c}-2\overrightarrow{a}.\overrightarrow{c}=2\mid \overrightarrow{a}{\mid }^2-\mid \overrightarrow{b}{\mid }^2$$$$\left(3\right)\overrightarrow{a}.\overrightarrow{b}=0$$$$then\overrightarrow{c}.(\overrightarrow{b}-\overrightarrow{a})=\overrightarrow{b}.\overrightarrow{c}-\overrightarrow{a}.\overrightarrow{c}$$$$=\mid \overrightarrow{a}{\mid }^2-\frac{1}{2}\mid \overrightarrow{b}{\mid }^2$$

Commented by nadovic last updated on 18/Jul/21

$$ThankyouverymuchSir$$

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