# Finding basis for the orthogonal complement [closed]

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Leucippus, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – John Hughes Apr 1 '18 at 0:06
• I got the answer [-3;1;0], [2;0;1] but the answer key to the exam has the answer shown above. – mymemesarespiciest Apr 1 '18 at 0:08
• You misunderstood the question. You computed a basis for $W$, but the question asked for a basis for the orthogonal complement of $W$. – amd Apr 1 '18 at 1:53

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 $$B=\{ (1,2,-3)\}$$

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 $$(1,2,-3)$$

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

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

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