線性代數之正交化(2題)

1. 解聯立方程組: (1) ab+d+fg=0, (2) ac+de+1/2f=0, (3)bc+e+1/2g=0, (4) a^2+d^2+f^2=1, (5) b^2+1+g^2=1, (6) c^2+e^2+1/4=1. (5)=> b=g=0; into (1) =>d=0; into (3)=>e=0; into (6)=>c=+-sqrt(3)/2. a and f satisfy the resulting (4) and (2), They are a^2+f^2=1 and +-sqrt(3)/2*a+1/2*f=0 => a=+-1(1/2), f=+-sqrt(3)/2, where +- signs can be selected arbitrarily as long as a*c and f are different in signs.

2. We first find a vector [a,b,c]^t so that together with [2 3 6]^t, and [3 -6 2]^t a basis of R^3 is formed. A choice of such vector is [6 2 -3]^t, which has norm=7. The projection of [49 49 49]^t on to [6 2 -3]^t is [30 10 -15]^t, which is obtaied by taking [49 49 49]^t{inner product}[6 2 -3]^t by the unit direction along [6 2 -3]^t. Thus [49 49 49]^t-[30 10 -15]^t would be totally on the subspace spanned by [2 3 6]^t, and [3 -6 2]^t, i.e. The answer is [19 39 64]^t.