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This is a snippet from my book.

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How did they get from |U|$^2$ = U • V = |U|•|V| |U|/|V| ?

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marked as duplicate by Najib Idrissi, Alex M., SchrodingersCat, Silvia Ghinassi, jameselmore Jan 28 '16 at 14:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @NajibIdrissi In this case, the question in the last line is about the algebraic manipulation $|\boldsymbol{u}|^2 = |\boldsymbol{u}|~|\boldsymbol{v}|~\frac{|\boldsymbol{u}|}{|\boldsymbol{v}|}$. $\endgroup$ – N. F. Taussig Jan 28 '16 at 13:32
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Well, $$|\vec{u}|^2=|\vec{u}|^2\frac{|\vec{v}|}{|\vec{v}|}=|\vec{u}||\vec{v}|\frac{|\vec{u}|}{|\vec{v}|}= |\vec{u}||\vec{v}|\text{cos}(\theta)\,\,,$$ since $|\vec{v}|/|\vec{v}|=1$.

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  • $\begingroup$ But isn't |U|^2 = U • U? So U•U = |U||U|? $\endgroup$ – Instinct May 16 '14 at 1:33
  • $\begingroup$ Yes, this doesn't contradict what I'm saying...I was explaining the third equality in your book's proof of the formula. $\vec{u} \cdot \vec{u} = |\vec{u}|^2$, then continue with what I have given. $\endgroup$ – afedder May 16 '14 at 1:40
  • $\begingroup$ I just didn't think U•U = |U||U|, thats why I thought it was a little confusing. But thank you. $\endgroup$ – Instinct May 16 '14 at 2:01
  • $\begingroup$ No problem, the statement you just said is true. @Instinct $\endgroup$ – afedder May 16 '14 at 2:30
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I dont understand the comments from your book, but I assume the question is how to prove the equivalence of the two definitions of dot product, $\mathbf{u}\cdot \mathbf{v}=u_1v_1+u_2v_2+u_3 v_3$, and the equation $\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta$. This is a consequence of the cosine law, if we take the triangle with sides $\mathbf{u},\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$ then the cosine law reads $$|\mathbf{u}-\mathbf{v}|^2=|\mathbf{u}|^2+|\mathbf{v}|^2-2|\mathbf{u}||\mathbf{v}|\cos \theta$$ using $$|\mathbf{u}-\mathbf{v}|^2==|\mathbf{u}|^2+|\mathbf{v}|^2 -2\mathbf{u}\cdot \mathbf{v}$$ we get the desired result.

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Under regular Euclidean geometry, the distance between two points $a=(x_1,y_1)$ and $b=(x_2,y_2)$ is defined as $$d(a,b)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$ Given a vector $\textbf{u}=(u_1,u_2,u_3)$, we know that $\textbf{u}=\sqrt{u_1^2+u_2^2+u_3^2}$. So from our definition of distance, we see that $\textbf{u}$ indicates the distance from the origin $(0,0,0)$ to $(a,b,c)$, or $\textbf{u}=d((0,0,0),\textbf{u})$. This relationship may help with the following explanation.

Using the regular notion of a dot product (i.e. $\textbf{u}\cdot\textbf{u}=u_1^2+u_2^2+u_3^2$), it is easy to check that $\textbf{||u||}^2=\textbf{u}\cdot\textbf{u}$ (see above). So from $0=\textbf{u}\cdot\textbf{v}-\textbf{u}\cdot\textbf{u}$, we have $\textbf{u}\cdot\textbf{v} =\textbf{u}\cdot\textbf{u}=\textbf{||u||}^2$. From there we multiply $\textbf{||u||}^2$ by $\frac{ \textbf{||v||}}{\textbf{||v||}}$. Finally, noticing that $\frac{\textbf{||u||}}{\textbf{||v||}} = cos\theta$, we have $$\textbf{u}\cdot\textbf{v}=||\textbf{u}||*||\textbf{v}||cos\theta.$$

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$\textbf{A really nice construction of this proof}$: Let $\textbf{A,B}$ be two vectors and $B$ $\neq O$. We seek a vector $P$ such that, $$(\textbf{A} -P) \cdot \textbf{B} = 0$$ and $P=c\textbf{B}$ for some number $c$. Suppose we can find such a number $c$, namely one satisfying, $$(A-cB) \cdot B = 0$$ We then get $$\textbf{A} \cdot \textbf{B} = c\textbf{B} \cdot \textbf{B}$$ and therefore $c=\frac{ \textbf{A} \cdot \textbf{B}}{\textbf{B} \cdot \textbf{B}}$. From this construction and use of what we know about planee geometry, $$\cos(\theta) = \frac{c||\textbf{B}||}{||\textbf{A}||}$$ and substituting in our value for $c$ we get, $$A \cdot B = ||\textbf{A}|| ||\textbf{B}|| \cos(\theta)$$ \

$\textbf{Picture for our Construction}$

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