Question

Let u=(-1,0,2), v=(3,1,2), w=(1,-2,-2) be vectors in standard position. Compute
1. The orthogonal complement of v

Answer 1

The orthogonal compliment of $v$ is

$v^{\perp}=\{ u\in V:<v,u>=0\}$

Let $u=(x,y,z)\neq 0\in v^{\perp}$

Then $<v,u>=3x+y+2z=0$

$..................(1)$

Now we have to find the nonzero solution of (1) .The free variable of equation (1) are x and z

(1) set $x=0,z=1$ to obtain the solution $v_1=(0,-2,1)$

(2) set $z=0,x=1$ to obtain the solution $v_2=(1,-3,0)$

The vector $v_1 \ and \ v_2$ form a basis for the solution space of the equation (1) and hence a basis for $v^{\perp}.$

$\therefore v^{\perp}=span\{(0,-2,1),(1,-3,0)\}$