[ad_1]
Topic
Give three distinctions I, A, REMOVE and real numbers k ≠ 1 satisfies \(\overrightarrow {IA} = k\overrightarrow {IB} \). Prove that with O is any point we have:
\(\overrightarrow {OI} = \left( {\frac{1}{{1 – k}}} \right)\overrightarrow {OA} – \left( {\frac{k}{{1 – k}} } \right)\overrightarrow {OB} \)
Solution method – See details
Split the vectors \(\overrightarrow {OA} ,\overrightarrow {OB} \) so that the vector \(\overrightarrow {OI} \) appears and combine the hypothesis to transform the right side
Detailed explanation
By assumption, \(\overrightarrow {IA} = k\overrightarrow {IB} \)
Consider the right side
We have:
VT = \(\left( {\frac{1}{{1 – k}}} \right)\overrightarrow {OA} – \left( {\frac{k}{{1 – k}}} \right) \overrightarrow {OB} = \left( {\frac{1}{{1 – k}}} \right)\left( {\overrightarrow {OI} + \overrightarrow {IA} } \right) – \left( { \frac{k}{{1 – k}}} \right)\left( {\overrightarrow {OI} + \overrightarrow {IB} } \right)\)
\( = \left( {\frac{1}{{1 – k}}} \right)\overrightarrow {OI} + \left( {\frac{1}{{1 – k}}} \right)\ overrightarrow {IA} – \left( {\frac{k}{{1 – k}}} \right)\overrightarrow {OI} – \left( {\frac{k}{{1 – k}}} \right )\overrightarrow {IB} \) \( = \left( {\frac{1}{{1 – k}} – \frac{k}{{1 – k}}} \right)\overrightarrow {OI} + \left( {\frac{1}{{1 – k}}} \right).k\overrightarrow {IB} – \left( {\frac{k}{{1 – k}}} \right)\overrightarrow {IB} \)\( = \overrightarrow {OI} + \left( {\frac{1}{{1 – k}}} \right).k\overrightarrow {IB} – \left( {\frac{k}{{1 – k}}} \right)\overrightarrow {IB} = \overrightarrow {OI} + \left( {\frac{k}{{1 – k}} – \frac{k}{{1 – k}}} \ right)\overrightarrow {IB} \) \( = \overrightarrow {OI} \) (DPCM)
[ad_2]
Source link net do edu
