Let $v_1, v_2, v_3$ be linearly independent vectors in $R^3$. Consider
$$v_3=v_2-\frac{(v_2)\cdot(v_1)}{\|v_1\|^2}v_1$$ Which of the following must be true? Note: $v_1\cdot v_2$ means the standard inner product of $v_1$ and $v_2$.
A. $v_3 \in span${$v_1,v_2$}
B. $v_3 \cdot v_1=0$
C. $v_3$ is orthogonal to $v_1$
D. $v_1$ is orthogonal to $v_2$ and $v_3$
E. $v_3 \in span${$v_1$}
F. $v_3 \cdot v_1 \neq0$

I know that $$\frac{(v_2)\cdot(v_1)}{\|v_1\|^2}v_1$$ stands for the projection of $v_1$ onto $v_2$. So then, the full equation $$v_3=v_2-\frac{(v_2)\cdot(v_1)}{\|v_1\|^2}v_1$$ is saying that $v_3$ is equal to the vector $v_2$ minus the projection of $v_1$. However, just because the three vectors are linearly independent, they don't have to be orthogonal. But, I don't have enough information to determine whether they are or not. If $v_1$ and $v_3$ are orthogonal, then the standard inner product is $0$. However, if they are orthogonal, doesn't that mean that the projection value is $0$ as well, and so $v_3=v_2$?

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    $\begingroup$ Well clearly $v_3 \in \mathrm{span}\{v_1, v_2\}$ right? $\endgroup$ – Alonso Delfín Aug 19 '17 at 1:21

Some thoughts to get you started.

Let $||v_i|| > 0$ and $$ v_3 = v_2 - \frac{v_1\cdot v_2}{||v_1||_2^2}v_1 = v_2 - av_1 $$ So $v_3$ is a linear combination of $v_1,v_2$.

Now consider $$v_3\cdot v_1 = v_2\cdot v_1 - a(v_1\cdot v_1)=v_2\cdot v_1 - \frac{v_1\cdot v_2}{||v_1||_2^2}||v_1||_2^2 = 0 $$ so $v_3\perp v_1$. Also $v_3\notin\text{span}\{v_1\} $ because $$ v_3 = bv_1 = v_2 - av_1 \implies v_2 = (a+b)v_1 $$ which implies $v_1,v_2$ are linearly dependent.


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