# Inner product with orthogonal complement

Let $$\mathbb{R^3}$$ be equipped with the inner product $$<,>$$ defined by setting

$$<\mathbf{u},\mathbf{v}>=2u_1v_1+5u_2v_2+3u_3v_3$$ for any pair of vectors $$\mathbf{u}=(u_1,u_2,u_3)$$ and $$\mathbf{v}=(v_1,v_2,v_3)$$. Which of the following subsets of $$\mathbb{R}^3$$ are bases of the orthogonal complement of the line $$L=span\left\{1,-1,1\right\}$$ in $$\mathbb{R^3}$$ relative to $$<,>$$?

i) $$\left\{(1,1,0),(0,1,1)\right\}$$

ii) $$\left\{(-2,4,8),(3,-6,-12)\right\}$$

iii) $$\left\{(3,6,-12),(1,2,-4)\right\}$$

iv) $$\left\{(-2,4,8),(1,1,1)\right\}$$

v) $$\left\{(-2,4,8),(0,0,0)\right\}$$

$$(A)$$ Just i.

$$(B)$$ Just iii.

$$(C)$$ Just iv.

$$(D)$$ ii and iv.

$$(E)$$ ii, iv and v.

My idea was to do the inner product with some arbitrary $$x$$ and $$y$$ such that:

$$\mathbf{x}=(x_1,x_2,x_3)$$

$$\mathbf{y}=(y_1,y_2,y_3)$$

Thus, $$<\mathbf{q},\mathbf{x}>=0$$ and $$<\mathbf{q},\mathbf{y}>=0$$ where

$$\mathbf{q}=(1,-1,1)$$

Thus:

$$<\mathbf{q},\mathbf{x}>=2x_1-5x_2+3x_3=0$$

$$<\mathbf{q},\mathbf{y}>=2y_1-5y_2+3y_3=0$$

Then we check each point in the answer choices and see which pair of $$x$$ and $$y$$ points work:

For instance in choice $$(i)$$, $$x=(1,1,0)$$ and $$y=(0,1,1)$$. Therefore:

$$2(1)-5(1)+3(0)=-3$$ (For $$x$$)

$$2(0)-5(1)+3(0)=-2$$ (For $$y$$)

Because this is not zero, this is not an orthogonal complement of the line $$L$$.

For choice $$(ii)$$:

$$2(-2)-5(4)+3(8)=0$$ (For $$x$$)

$$2(3)-5(-6)+3(-12)=0$$ (For $$y$$)

Because this is zero, this is an orthogonal complement of the line $$L$$.

Similar calculations for the other points yield that only $$ii,iv,v$$ work and thus the answer is $$(E)$$. However, it says the correct answer is $$(C)$$. Why is that?

The basis of $$L^\perp$$ consists of $$(2)$$ linearly independent vectors. The vectors in $$(ii),(v)$$ are linearly dependent, and can't form the basis of $$L^\perp$$.
As an additional comment on your method of solving, you don't need to take two vectors $$\bf x,y$$. Just take a general vector $$\mathbf x\in L^\perp$$, and the condition you get must be satisfied by all vectors in $$L^\perp$$. To then find the basis of $$L^\perp$$, you need to see which option contains the required number of linearly independent vectors that satisfy the condition.
• Is the dimension $2$ because of the fact you need two sets points to span a line? In addition, why is $iv$ linearly dependent? I can't write vector given as a linear combination of the other. That's impossible. – Future Math person Dec 21 '18 at 21:13
• Any set of vectors containing the zero vector is linearly dependent. For linear dependence of vectors $v_1,v_2$;$$c_1v_1+c_2v_2=0$$ must have a non-trivial solution for $c_1,c_2$. Take $v_2=(0,0,0),c_1=0$ and $c_2=k\ne0$. – Shubham Johri Dec 21 '18 at 21:18
• But that's choice $(v)$ not $(iv)$. – Future Math person Dec 21 '18 at 21:18
• Anywho, the dimension of $L^\perp$ (which is a plane, not a line) is $2$ because any vector $v=(x_1,x_2,x_3)\in L^\perp$ satisfies $2x_1-5x_2+3x_3=0$. So $(x_1,x_2,x_3)=(x_1,\frac{2x_1+3x_3}5,x_3)=x_1(1,2/5,0)+x_3(0,3/5,1)$, that is, $L^\perp$ is spanned by two linearly independent vectors – Shubham Johri Dec 21 '18 at 21:24