Question

Vă rog.. nu prea mă descurc la vectori.. cum rezolv daca 2 vectori sunt coliniari?

Answer 1

(xN-xM)/(yN-yM)=(xP-xM)/(yP-yM)

(n-1)/(2n-1)=(3-n)/(5-n)

(5-n) (n-1)=(2n-1)(3-n)

-n²+6n-5=-2n²+7n-3

n²-n-2=0

Rez\olvand cuΔ obtinem

n1=-1 si n2=2

Extra

pt n=-1 punctele sunt (1;-1); (-1;3) ;(-2;5) pe drepta y=-2x+1, deci da, coliniare

pt n=2 punctele sumt (1;2) (2;3);(4;5) pe dreapta y=x+1, deci da, coliniare

deci BINE REZOLVAT

Answer 2

$\it \overrightarrow{MN} = (x_N-x_M)\overrightarrow{i} + (y_N-y_M)\overrightarrow{j} = (n-1)\overrightarrow{i} + (3-n)\overrightarrow{j} \\ \\ \\\overrightarrow{MP} = (x_P-x_M)\overrightarrow{i} + (y_P-y_M)\overrightarrow{j} = (2n-1)\overrightarrow{i} + (5-n)\overrightarrow{j} \\ \\ \\ \overrightarrow{MN},\ \overrightarrow{MP} \ coliniari \Leftrightarrow \dfrac{n-1}{2n-1} = \dfrac{3-n}{5-n} \Leftrightarrow (n-1)(5-n)=(2n-1)(3-n)\\ \\ \\ \Leftrightarrow 5n-n^2-5+n = 6n-2n^2-3+n\Leftrightarrow$

$\it \Leftrightarrow 5n-n^2-5+n - 6n+2n^2+3-n=0 \Leftrightarrow n^2-n-2=0\Leftrightarrow\\ \\ \\\Leftrightarrow n^2+n-2n-2=0\Leftrightarrow n(n+1)-2(n+1)=0\Leftrightarrow(n+1)(n-2)=0 \Leftrightarrow\\ \\ \\\Leftrightarrow \begin{cases} \it n+1=0 \Rightarrow n=-1\\ \\sau\\ \\\it n-2=0 \Rightarrow n=2\end{cases}$