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Q.

The P.V's of A, B and $\bar{a}, \bar{b}$ respectively. The P.V of 'C' is $\frac{3 \bar{a}}{4} + \frac{\bar{b}}{2}$ then 'C' lies

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a

outside $∆ OAB$ but inside $∠ OAB$

b

outside $∆ OAB$ but inside $∠ OBA$

c

inside $∠ OAB$

answer is A.

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Detailed Solution

$\bar{O A} = \bar{a}, \bar{O B} = \bar{b} \bar{O C} = \frac{3 \bar{a}}{4} + \frac{\bar{b}}{2} 4 \bar{O C} = 3 \bar{O A} + 2 \bar{O B}$

inside $OAB$

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