Question

Decomponha W = (1,0,3) como a soma de dois vetores W1 e W2, com W1 paralelo ao vetor (0,1,3) e W2 ortogonal a este último.

Answer 1

Resposta:

W1=k*(0,1,3)

W2=(a,b,c)

W1 e W2 são ortogonais

W1.W2 = 0   ...produto escalar

k*(0,1,3).(a,b,c)=0

k*b+3k*c=0

b=-3c

W2=(a,-3c,c) =a*(1,0,0) +c*(0,-3,1)

W=q*W1+t*W2

W=k*(0,1,3) + a*(1,0,0) +c*(0,-3,1)

(1,0,3)=k*(0,1,3) + a*(1,0,0) +c*(0,-3,1)

1=a

0=k-3c ==>k=3c=9/10

3=3k+c ==>3=9c+c ==>c=3/10

W1=(9/10)*(0,1,3)=(0,9/10,27/10)

W2=1*(1,0,0) +(3/10)*(0,-3,1) =(1,-9/10,3/10)

(1,0,3)= (0, 9/10, 27/10) + (1 ,-9/10 , 3/10) é a decomposição

Answer 2

$W1 = \frac{W*V}{V^{2} } *V$

$W1 = \frac{(1,0,3)*(0,1,3)}{(0,1,3)^{2} } *(0,1,3)$

$W1 = \frac{1*0+0*1+3*3}{0^{2}+1^{2}+3^{2} } *(0,1,3)$

$W1 = \frac{0+0+9}{0+1+9} *(0,1,3)$

$W1 = \frac{9}{10} *(0,1,3)$

$W1 = (0, \frac{9}{10}, \frac{27}{10} )$

$W2 = W - W1$

$W2 = (1,0,3) - (0, \frac{9}{10},\frac{27}{10} )$

$W2 = (1, -\frac{9}{10} , \frac{3}{10} )$

W1 = (0, 9/10, 27/10) e W2 = (1, - 9/10, 3/10)