Question #219100
For the subspace W= the plane 3x-y+4z-0, of R
1
Expert's answer
2021-07-22T04:50:52-0400

The normal to the plane has the direction (3,1,4).(3, -1, 4).

Therefore, the basis must consist of two linearly independent vectors that are orthogonal to it.


3a1b1+4c1=03a_1-b_1+4c_1=0

3a2b2+4c2=03a_2-b_2+4c_2=0

We consider the set {(0,4,1),(1,3,0)}.\{(0, 4, 1), (1, 3, 0)\}.

We can see that for any scalar k,k,


(0,4,1)k(1,3,0)(0, 4, 1)\not=k (1, 3, 0)

Therefore, they are linearly independent.

Also,


(0,4,1)(3,1,4)=0(0, 4, 1)\cdot(3, -1, 4)=0

(1,3,0)(3,1,4)=0(1, 3, 0)\cdot(3, -1, 4)=0

Therefore, the basis is {(0,4,1),(1,3,0)}.\{(0, 4, 1), (1, 3, 0)\}.



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