Solve Lesson 75 Page 107 Math 10 – Kite >

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)