A vector orthogonal to two others (in 3-d space) can be obtained by the cross product.
A=
x y z
1 5 3
1 -5 1
Evaluate A
(5+15,-1*(1)+1*3,1(-5)-1(5))
=(20,2,10)
=(5,1/2,-5/2)
Check:
(5,1/2,-5/2).(1,5,3)=0
(5,1/2,-5/2).(1,-5,1)=0
So solution is good.
Answer: b=1/2, c=-5/2
Alternatively, by orthogonal conditions,
multiply (5,b,c) by (1,5,3) = (3c+5b+5) => 3c+5b+5=0.....(1)
multiply (5,b,c) by (1,-5,1) = (c-5b+5) => c-5b+5=0....(2)
Solve (1) & (2) for b and c
(1)+(2) => 4c+10=0 => c=-5/2
Substitute c in 2
-5/2-5b+5=0 => b=1/2
So we get the same results.
