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Vector Addition

The diagram s...

Question

The diagram shows a scalene triangle, $A B C$ , with vectors $- - \to A B =^a$ and $- - \to A C = 2^b$ . The point $D$ is positioned on the line $B C$ such that $B D : D C = 3 : 2$ . The point $E$ is positioned on the line $A D$ such that $A E : E D = 1 : 2$ . The expression for the vector $- - \to A E$ in terms of $^a$ and $^b$ is

A

(\frac{2^b}{15} + \frac{2^a}{5})

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B

(\frac{2^b}{5} + \frac{4^a}{15})

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C

(\frac{2^b}{5} + \frac{2^a}{15})

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D

(\frac{3^b}{5} + \frac{2^a}{15})

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Solution

The correct option is C $(\frac{2^b}{5} + \frac{2^a}{15})$
First we have vector $- - \to B C$ as
$- - \to B C = - - \to A C - - - \to A B = 2^b -^a$
We have
$- - \to B D = \frac{3}{5} - - \to B C = \frac{3}{5} (2^b -^a) = (\frac{6^b}{5} - \frac{3^a}{5})$
Also, $- - \to A D = - - \to A B + - - \to B D$
$\Rightarrow - - \to A D =^a + (\frac{6^b}{5} - \frac{3^a}{5}) = (\frac{6^b}{5} + \frac{2^a}{5})$
Thus, $- - \to A E = \frac{1}{3} (- - \to A D) = \frac{1}{3} (\frac{6^b}{5} + \frac{2^a}{5})$
$\Rightarrow - - \to A E = (\frac{2^b}{5} + \frac{2^a}{15})$

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