Proof involving projection

Question

I need to prove the following:

If $\vec{A}$ and $\vec{B}$ are two vectors different from $\vec{0}$. Proof $\vec{A} - c \vec{B} $ is orthogonal to $\vec{B}$ if $ c = \frac{\vec{A} · \vec{B} }{||\vec{B} ||^2}$.

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I did some attempts and I got that given $c =\frac{\vec{A} · \vec{B} }{||\vec{B} ||^2}$ then $cB$ is the orthogonal projection of $\vec{A}$ onto $\vec{B}$. Thus, $\vec{A} - c \vec{B} $ is the vector that connects the ends of $\vec{A}$ and it's projection on $\vec{B}$. Therefore, $(\vec{A} - c \vec{B}) · \vec{B} = 0$ and it's orthogonal.

I was told that I need it to be more formal but I don't know how to make it more formal. Thanks.

Answer 1

When you say things like 'is the vector that connects the ends of $\vec{A}$ and its projection on $\vec{B}$', you are interpreting the result and thinking about the usual Euclidian space. Note that there is a certain gap here, because we don't know precisely what 'the end' of a vector is. Also, this result can be stated in a more general context, so it is useful to think in a more abstract way instead of interpreting it geometrically.

What about this approach: let $\vec{v} = \vec{A}-c\vec{B}$. Then, you have: $$ \vec{v}\cdot \vec{B} = (\vec{A}-c\vec{B})\cdot \vec{B} = \vec{A}\cdot \vec{B}-c\vec{B}\cdot \vec{B}$$ Now, remember that $\vec{B}\cdot \vec{B} = ||\vec{B}||^{2}$. Take $c = \frac{\vec{A}\cdot \vec{B}}{||B||^{2}}$. Then: $$\vec{v}\cdot \vec{B} = \vec{A}\cdot \vec{B}-\frac{\vec{A}\cdot \vec{B}}{||\vec{B}||^{2}}||\vec{B}||^{2} = \vec{A}\cdot \vec{B} -\vec{A}\cdot\vec{B} = 0 $$ so that $\vec{v}= \vec{A}-c\vec{B}$ with $c = \frac{\vec{A}\cdot\vec{B}}{||\vec{B}||^{2}}$ is orthogonal to $\vec{B}$.

Answer 2

To do this more formally, you can write an algebraic proof. Thus, \begin{align*}(\overrightarrow{A} - c\overrightarrow{B})\cdot\overrightarrow{B} &= \overrightarrow{A}\cdot\overrightarrow{B} - \left(c\overrightarrow{B}\right)\cdot\overrightarrow{B} \qquad \text{(by distributivity of the dot product)} \\ &= \overrightarrow{A}\cdot\overrightarrow{B} -c\left(\overrightarrow{B}\cdot\overrightarrow{B}\right) \qquad \text{(by linearity of the dot product)} \\ &= \overrightarrow{A}\cdot\overrightarrow{B} -c\left\lVert \overrightarrow{B}\right\rVert^2 \qquad \text{(by the definition of $\left\lVert .\right\rVert$)} \\ &= \overrightarrow{A}\cdot\overrightarrow{B} -\frac{\overrightarrow{A}\cdot\overrightarrow{B}}{\left\lVert \overrightarrow{B}\right\rVert^2}\left\lVert \overrightarrow{B}\right\rVert^2 \\ &= 0\end{align*} as required. $\qquad \rule{0.7em}{0.7em}$