(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.

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

By definition, this means we want

and we can deduce

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 as

In other words, we just proved that any vector of 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

5050 i think

step-by-step explanation: