Solve Lesson 63 Page 106 SBT Math 10 – Kite>

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Topic

Give four points A, REMOVE, OLD, EASY. Prove \(\overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {AC} .\overrightarrow {DB} + \overrightarrow {AD} .\overrightarrow {BC} = 0\)

Solution method – See details Split the vector and return the original common vectors (originalA

)

Detailed explanation

Transform the left side

We have:

\(\overrightarrow {AB} .\overrightarrow {CD} + \overrightarrow {AC} .\overrightarrow {DB} + \overrightarrow {AD} .\overrightarrow {BC} = \)\(\overrightarrow {AB} .\left ( {\overrightarrow {AD} – \overrightarrow {AC} } \right) + \overrightarrow {AC} .\left( {\overrightarrow {AB} – \overrightarrow {AD} } \right) + \overrightarrow {AD} . \left( {\overrightarrow {AC} – \overrightarrow {AB} } \right)\)\( = \overrightarrow {AB} .\overrightarrow {AD} – \overrightarrow {AB} .\overrightarrow {AC} + \overrightarrow {AB} .\overrightarrow {AC} – \overrightarrow {AC} .\overrightarrow {AD} + \overrightarrow {AC} .\overrightarrow {AD} – \overrightarrow {AB} .\overrightarrow {AD} = 0\) = VP(PCM)

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