Answer to Question #219100 in Linear Algebra for Shreya

Question

For the subspace W= the plane 3x-y+4z-0, of R

Answer 1

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

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

$$3a_1-b_1+4c_1=0$$

$$3a_2-b_2+4c_2=0$$

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

We can see that for any scalar $k,$

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

Therefore, they are linearly independent.

Also,

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

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

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