Find the orthogonal complement of W and give a basis for W={[xyz]:x=12t, y=−12, z=2t}

So we need to find two linearly independent vectors orthogonal to[12−122][1−14](or to avoid fractions). To do so, we can just use the definition of orthogonality:(x,y,z)⋅(1,−1,4)=x−y+4z=0Let's let x=1 and y=0. Then we see that z=−14. So[10−14]is in W. Now let x=0 and y=1. Then z=14. So[0114]is also in W. ThusW={[10−14],[0114]}where{[10−14],[0114]}Is a basis for W

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cleritere39

Answered question

2021-11-21

Find the orthogonal complement of W and give a basis for

W = {[\begin{matrix} x \\ y \\ z \end{matrix}] : x = \frac{1}{2} t, y = - \frac{1}{2}, z = 2 t}

Answer & Explanation

Walker Funk

Beginner2021-11-22Added 13 answers

So we need to find two linearly independent vectors orthogonal to

[\begin{matrix} \frac{1}{2} \\ - \frac{1}{2} \\ 2 \end{matrix}]

[\begin{matrix} 1 \\ - 1 \\ 4 \end{matrix}]

(or to avoid fractions). To do so, we can just use the definition of orthogonality:

(x, y, z) \cdot (1, - 1, 4) = x - y + 4 z = 0

Let's let x=1 and y=0. Then we see that

z = - \frac{1}{4}

. So

[\begin{matrix} 1 \\ 0 \\ - \frac{1}{4} \end{matrix}]

is in W. Now let x=0 and y=1. Then

z = \frac{1}{4}

. So

[\begin{matrix} 0 \\ 1 \\ \frac{1}{4} \end{matrix}]

is also in W. Thus

W = {[\begin{matrix} 1 \\ 0 \\ - \frac{1}{4} \end{matrix}], [\begin{matrix} 0 \\ 1 \\ \frac{1}{4} \end{matrix}]}

where

{[\begin{matrix} 1 \\ 0 \\ - \frac{1}{4} \end{matrix}], [\begin{matrix} 0 \\ 1 \\ \frac{1}{4} \end{matrix}]}

Is a basis for W

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