To find the surfaces of a+b, b+c and c+a, the inverse method can be used, assuming that the three surfaces are * * *.

The * * * plane of three vectors, namely k 1 (a+b)+K2 (b+c)+K3 (c+a) = 0, and at least1in k 65438, k2 and k3 is not zero.

The three vectors are not * * * planes, that is, k1(a+b)+k2 (b+c)+k3 (c+a) = 0. At this time, K 1, K2 and K3 can only all be 0.

A, b and c are cardinality, so they must not be * * * planes, that is, x 1a+x2b+x3c=0, x 1, x2 and x3 can only be 0.

Because k 1, k2 and k3 are all symmetrical, the problem is that the default k 1 is not 0.

Then divide the two ends of the equation by k 1 and get (a+b)+k2/k1(b+c)+k3/k1(c+a) = 0.

Let μ =-k3/k 1 and λ=-k2/k 1, which is the representation in the answer.

If you want to know whether k 1, k2 and k3 are non-zero, you need to calculate.

K1(a+b)+k2 (b+c)+k3 (c+a) = 0 or (a+b)= λ(b+c) +μ(c+a).

That is, (k1+k3) a+(k1+k2) b+(k2+k3) c = 0.

Or (1-μ) a+(1-λ) b-(λ+μ) c = 0.

Because a, b and c are cardinality, then (k1+k3) = (k1+k2) = (k2+k3) = 0 ①.

Or (1-μ) = (1-λ) = (λ+μ) = 0 ②.

Only k 1=k2=k3=0 can be found.

Or μ=-k3/k 1 =-0/0, λ=-k2/k 1=-0/0 has no solution.

So three vectors are not * * * planes.

Newman Hero 2065438+March 28th, 2005 10:05: 13

I hope it will help you and I hope it will be adopted.