Question

Sean los vectores:

a→=(2;−1)yb→=(−3;2)

Halle: Proy12a→5b→+Proya→3b→

Answer 1

Al realizar la operación con vectores se obtiene:

Proy 12a·5b + Proy a·3b = (-128/5; 64/5)

Explicación paso a paso:

Datos;

a = (2; -1)

b = (-3; 2)

Halle: Proy 12a·5b + Proy a·3b

Producto de un escalar por un vector;

12a = 12(2;-1)

12a = (24; -12)

5b = 5(-3; 2)

5b = (-15; 10)

3b = 3(-3; 2)

3b = (-9; 6)

Formula de proyección de dos vectores;

$Proy_v u =(\frac{u.v}{|v|^{2} } ).v$

$Pory_{12a}5b =(\frac{12a.5b}{|12a|^{2} } ).12a$

12a·5b = (24)(-15)+(-12)(10) = -360-120 = -480

|12a|² = (24)² + (-12)²  = 720

Sustituir;

$Pory_{12a}5b =(\frac{-480}{720} ).(24;-12)\\Pory_{12a}5b =(-16;8)$

$Pory_{a}3b =(\frac{a.3b}{|a|^{2} } ).a$

a = (2;-1)

a·3b = (2)(-9) + (-1)(6) = -18-6 = -24

|a|² = (2)² + (-1)² = 4+1 = 5

Sustituir;

$Pory_{a}3b =(\frac{-24}{|5} ).(2;-1)\\Pory_{a}3b =(-\frac{48}{5};\frac{24}{5})$

Proy 12a·5b + Proy a·3b = (-16 -48/5; 8+24/5)

Proy 12a·5b + Proy a·3b = (-128/5; 64/5)