Vector

Geometry Level 3

The coordinates of A , B A, B and C C are ( 1 , 2 ) , ( 7 , 1 ) (1,2), (7,1) and ( 15 , 9 ) (15,9) , respectively. O O is the origin and O C = h O A + k O B \overrightarrow{OC}\,=\,h\overrightarrow{OA}\,+\,k\overrightarrow{OB} , where h h and k k are constants. If h + k h+k can be expressed as m n \dfrac{m}{n} , where m m and n n are coprime positive integers , find m + n m+n .


The answer is 82.

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1 solution

Armain Labeeb
Jul 16, 2016

Let the position vectors of A A , B B and C C be a a , b b and c c respectively.

c = h a + k b ( 15 9 ) = h ( 1 2 ) + k ( 7 1 ) = ( h + 7 k 2 h + k ) h + 7 k = 15 ( 1 ) 2 h + k = 9 ( 2 ) 2 h + 14 k = 30 ( 3 ) ( 1 ) × 2 13 k = 21 ( 3 ) ( 2 ) k = 21 13 h = 15 7 ( 21 13 ) = 48 13 Substituting the value of k into (1) h + k = 48 + 21 13 = 69 13 m n = 69 13 m + n = 69 + 13 = 82 \begin{aligned} c & =ha+kb & \\ \left( \begin{matrix} 15 \\ 9 \end{matrix} \right) & =h\left( \begin{matrix} 1 \\ 2 \end{matrix} \right) +k\left( \begin{matrix} 7 \\ 1 \end{matrix} \right) \, =\, \left( \begin{matrix} h+7k \\ 2h+k \end{matrix} \right) & \\ h\, +\, 7k\, & =\, 15 \quad \quad \longrightarrow (1) & \\ 2h\, +\, k\, & =\, 9\quad \quad \, \, \, \longrightarrow (2) & \\ 2h\, +\, 14k\, & =\, 30\quad \quad \longrightarrow (3) & (1) \, \times \, 2 \\ 13k\, & =\, 21 & (3)\,-\,(2) \\ k\, & =\, \frac { 21 }{ 13 } & \\ h\, & =15-7\left( \frac { 21 }{ 13 } \right) \, =\, \frac { 48 }{ 13 } & \text{Substituting the value of k into (1)} \\ \therefore \quad h\, +\, k\, & =\, \frac { 48+21 }{ 13 } \, \, =\, \frac { 69 }{ 13 } & \\ \Longrightarrow \quad \frac { m }{ n } \, & =\, \frac { 69 }{ 13 } & \\ \therefore \quad \quad \quad \quad \quad m+n\, & =\,69+13\\ & =\,\boxed{82} \end{aligned}

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