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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ – Future Math person Dec 21 '18 at 21:13
  • $\begingroup$ 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 $c_2=k\ne0$. $\endgroup$ – Shubham Johri Dec 21 '18 at 21:18
  • $\begingroup$ But that's choice $(v)$ not $(iv)$. $\endgroup$ – Future Math person Dec 21 '18 at 21:18
  • $\begingroup$ Oh whoops...see the changed answer $\endgroup$ – Shubham Johri Dec 21 '18 at 21:19
  • $\begingroup$ 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 $\endgroup$ – Shubham Johri Dec 21 '18 at 21:24

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