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The following excerpt can be found in Serge Lang's Introduction to Linear Algebra. I am trying to understand mathematically why the vector $\mathbf{A}- c\mathbf{B}$ is perpendicular to the vector $c\mathbf{B}$. I suppose there would be a simple mathematical explanation behind this, but I haven't been able to find any. I have tried taking the dot product of $\mathbf{A} - c\mathbf{B}$ and $c\mathbf{B}$ to equal $0$, but I cannot find any proof as to why this dot product would have to equal $0$.

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    $\begingroup$ It won't be true for an arbitrary value of $c$ .Presumably Lang goes on to calculate what value of $c$ makes it true? $\endgroup$
    – user169852
    Jul 3 '20 at 1:11
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    $\begingroup$ Can you provide the complete question or the page no. of your context from the book which you have taken. $\endgroup$
    – SarGe
    Jul 3 '20 at 1:51
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    $\begingroup$ From the part of the text you show, it seems like he is defining the projection to be the unique vector $P$ which is parallel to $B$ and such that $A-P$ is perpendicular to $B$. At least secretly. $\endgroup$
    – Keshav
    Jul 3 '20 at 2:00
  • $\begingroup$ @Doubtnut The diagram was on page 23 of his second edition. $\endgroup$
    – Eli Jones
    Jul 3 '20 at 2:06
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    $\begingroup$ I've edited the question by posting image of the complete section which might be helpful for other users to understand the scenario. $\endgroup$
    – SarGe
    Jul 3 '20 at 2:17
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As @Bungo has mentioned, it is not true for an arbitrary value $c\in\textbf{F}$. It just states the projection of $A$ lies in the direction $B$. More precisely, in order to find $c$, it has to satisfy the following relation: \begin{align*} \langle A-cB,cB\rangle = 0 & \Longleftrightarrow \langle A,cB\rangle - \langle cB,cB\rangle = 0\\\\ & \Longleftrightarrow \overline{c}\langle A,B\rangle - c\overline{c}\langle B,B\rangle = 0 \end{align*}

If $B\neq 0$ and $c\neq 0$, it results that \begin{align*} \langle A,B\rangle - c\langle B,B\rangle = 0 \Longleftrightarrow c = \frac{\langle A,B\rangle}{\langle B,B\rangle} \end{align*} and we are done.

Hopefully it helps.

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