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Suppose $W$ consists of all vectors $(x,y,z)$ such that $x + 2y - 3z = 0$. Which of the following is a basis for the orthogonal complement?

The answer is $(1,3,-2)$ but I don't understand that at all. How do I get to that answer?

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closed as off-topic by Shailesh, Leucippus, Alexander Gruber Apr 1 '18 at 3:42

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    $\begingroup$ Do you notice anything about the coefficients in the linear equation? Also: what do you know about orthogonality? Finally: the answer you wrote down is actually wrong; if that's the answer in the answer book, someone made a typo. $\endgroup$ – John Hughes Apr 1 '18 at 0:06
  • $\begingroup$ I got the answer [-3;1;0], [2;0;1] but the answer key to the exam has the answer shown above. $\endgroup$ – mymemesarespiciest Apr 1 '18 at 0:08
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    $\begingroup$ You misunderstood the question. You computed a basis for $W$, but the question asked for a basis for the orthogonal complement of $W$. $\endgroup$ – amd Apr 1 '18 at 1:53
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You have a plane $$ x + 2y - 3z = 0$$

which is a two dimensional subspace of your three dimensional space.

The orthogonal complement is a one dimensional subspace which is apanned by a vector perpendicular to the plane.

The normal vector to your plane $$N= (1,2,-3) $$is such a vector.

Thus the basis for yor the orthogonal complement is $$ B=\{ (1,2,-3)\}$$

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Because linearity of inner product, orthogonal complement of a finite dimensional subespace can be obtained by its basis. A basis B for W is $B=(-2,1,0),(3,0,1)$

Then If $(x,y,z) \in W^{\perp}$ is orthogonal to every vector in B $$ ((x,y,z),(-2,1,0))=0$$ $$ ((x,y,z),(3,0,1))=0$$ Therefore $$ -2x+y=0 $$ $$ 3x+z=0 $$ It means a basis for orthogonal complement is $$ (1,2,-3) $$

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A basis for this plane is any pair of linearly independent vectors that start at a point in it and end at another point in it. So, consider two points in the plane $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$, then a vector in the plane would be $(x_2-x_1,y_2-y_1,z_2-z_1).$

Since both points are the plane, we have that

$$z_1 = \frac{x_1+2y_1}{3}\quad\text{and}\quad z_2 = \frac{x_2+2y_2}{3}$$

so that

$$z_2-z_1 = \frac{x_2-x_1+2(y_2-y_1)}{3}.$$

This is the only restriction that the vectors must satisfy. We need to pick two linearly independent vectors that satisfy this restriction. A simple way to do that is to set $x_2-x_1=1$ and $y_2-y_1=0$ for the first one, so that

$$z_2-z_1 = \frac{1+2(0)}{3}=\frac{1}{3}.$$

and $x_2-x_1=0$ and $y_2-y_1=1$ for the second one, so that

$$z_2-z_1 = \frac{0+2(1)}{3}=\frac{2}{3}.$$

It follows that a basis for this plane is

$$ \left(1,0,\frac{1}{3}\right)\quad\text{and}\quad\left(0,1,\frac{2}{3}\right).$$

To make a vector, $(a,b,c)$, orthogonal to these two we can, then, use the fact that

$$(a,b,c)\times \left(1,0,\frac{1}{3}\right)=0\;\Rightarrow\; a+\frac{c}{3}=0\;\Rightarrow\; a=-\frac{c}{3},$$

and

$$(a,b,c)\times \left(0,1,\frac{2}{3}\right)=0\;\Rightarrow\; b+\frac{2c}{3}=0\;\Rightarrow\; b=-\frac{2c}{3}.$$

You have one degree of freedom so just set $c=-3$, then $a=1$ and $b=2$. A vector orthogonal to the plane is then

$$(a,b,c)=(1,2,-3).$$

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  • $\begingroup$ Thank you. I noticed it as soon as I finished. I fixed the problem. $\endgroup$ – mzp Apr 1 '18 at 0:44

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