Question

(1 pt) let l be the line given by the span of \left[\begin{array}{c} 2\cr -6\cr 9 \end{array}\right] in \mathbb{r}^3. find a basis for the orthogonal complement l^{\bot} of l.

Answer 1

Two vectors are orthogonal if their scalar product is zero. Also, since we have a subspace of dimension 1 (the line) in a space of dimension 3 (R^3), the orthogonal complement has dimension 2.

This means that we're looking for a basis of the 2-dimensional subspace of R^3 defined by the property

$l^{\bot} = \left\{(x,y,z) \in \mathbb{R}^3\ :\ (2, -6, 9)\cdot (x,y,z) = 0\right\}$

By definition, this means we want

$2x-6y+9z=0$

and we can deduce

$2x-6y+9z=0 \iff z = \dfrac{-2x+6y}{9}$

We obviously have 2 degree of freedom here (we're building a base of a 2-dimensional space): we can write a generic vector of $l^{\bot}$ as

$\left(x, y, \dfrac{-2x+6y}{9}\right) = \left(x, 0, \dfrac{-2x}{9}\right) + \left(0, y, \dfrac{6y}{9}\right) = x\left(1,0,\dfrac{2}{9}\right) + y\left(0, 1, \dfrac{2}{3}\right)$

In other words, we just proved that any vector of $l^{\bot}$ can be written as a certain linear combination of vectors (1,0,-2/9) and (0,1,2/3), which means that they are a base of $l^{\bot}$