Question

Decomponha o vetor -»w = (1, 0, 3) como soma de dois vetores -»w1 e
-»w2, com (-»w1,(1, 1, 1),(−1, 1, 2)) linearmente dependentes e -»w2 ortogonal a estes dois últimos.

Answer 1

W = W1 + W2 --> W1 = W - W2

W2 ⊥ (1,1,1)

W2 ⊥ (-1,1,2)

(1,1,1) ∧ (-1,1,2)

$\left[\begin{array}{ccc}i&j&k\\1&1&1\\-1&1&2\end{array}\right]$

i*1*2+j*1*(-1)+k*1*1-((-1)*ik+1*1i+2*1j)

2i-j+k-(-k+i+2j)

2i-j+k+k-i-2j

i-3j+2k (1,-3,2)

W2 = K*(1,1,1) ∧ (-1,1,2)

W2 = (K,-3K,2K)

W1 = (1-K,3K,3-2K)

{W1,(1,1,1),(-1,1,2)} LD

$\left[\begin{array}{ccc}1-K&3K&3-2K\\1&1&1\\-1&1&2\end{array}\right] =0$

$(1-K)*1*2+(3K)*1*(-1)+(3-2K)*1*1-((-1)*1(3-2K)+1*(1-K)+2*1*3K) = 0$

$2-2K-3K=3-2K-(-(3-2K)+1-K+6K) = 0$

$2-2K-3K+3K-2K-(-3K+2K+1-K+6K)=0$

$2-2K-3K+3-2K-(-2+7K)=0$

$2-2K-3K+3-2K+2-7K=0$

$7-14K=0$

$-14K=-7 .(-1)$

$14K = 7$

$K = \frac{7}{14}$

$K=\frac{1}{2}$

W1 =

$1-\frac{1}{2} = \frac{2-1}{2} =\frac{1}{2}$

$\frac{3}{2}$

$3-\frac{2}{2} =3-1=2$

W2 =

$\frac{1}{2}$

$-\frac{3}{2}$

$\frac{2}{2} =1$

W1 = (1/2, 3/2, 2) e W2 = (1/2, -3/2, 1)