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I know the terminology is very weird and probably misled many people form what I’m actually asking, but I don’t know any better way to describe it...

Why is the dot product of $\ b_1 $ and $\ b_2 $ equal to equation below? How did they even get those values?

Can someone guide me through the steps to get the answer?

$\vec b_1 = \vec c_1 \cos \theta + \vec c_2 \sin \theta$

$\vec b_2 = \vec c_2 \cos \theta - \vec c_1 \sin \theta$

$\vec b_1 \cdot \vec b_2 = -(\vec c_1^2 - \vec c_2^2)\sin \theta \cos \theta + \vec c_1 \cdot \vec c_2(\cos^2 \theta - \sin^2 \theta) = 0$

the equation in an image

[edit 1] wrote out the equation in MathJax

[edit 2] forgot to mention that $ b_1 $ and $b_2$ are assumed to be perpendicular

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    $\begingroup$ Dot product is distributive over addition, so you can just expand the product. $\endgroup$ Commented Dec 27, 2019 at 1:36
  • $\begingroup$ Please make your question visible. You cannot "convert" that link into mathjax but you can type it in mathjax. $\endgroup$
    – Lee Mosher
    Commented Dec 27, 2019 at 1:36
  • $\begingroup$ MathJax quick reference $\endgroup$
    – amd
    Commented Dec 27, 2019 at 1:45
  • $\begingroup$ @eyeballforg I still don't understand what the simplified version of the dot product would be. I don't see any components of the vectors, can you please point them out? $\endgroup$
    – Sceptual
    Commented Dec 27, 2019 at 2:03
  • $\begingroup$ @Hasith when eyeballfrog says “expand the product,” he doesn’t mean that you should write the product out in terms of its components $\endgroup$ Commented Dec 27, 2019 at 2:13

1 Answer 1

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Given $\vec b_1$ and $\vec b_2$ are perpendicular and their dot product is 0, we have: $$ \vec b_1\cdot\vec b_2=(\vec c_1\cos\theta+\vec c_2\sin\theta)\cdot(\vec c_2\cos\theta -\vec c_1\sin\theta)=0$$ $$ \Rightarrow \vec c_1\cdot\vec c_2 \cos^2\theta - |\vec c_1|^2 \sin\theta\cos\theta +|\vec c_2|^2\sin\theta\cos\theta-\vec c_1\cdot\vec c_2 \sin^2\theta=0$$ $$\Rightarrow \vec c_1\cdot\vec c_2(\cos^2\theta-\sin^2\theta)-(|\vec c_1|^2-|\vec c_2|^2)\sin\theta\cos\theta=0$$ Notice that $\vec c_1 \cdot \vec c_2 $ is just a scalar quantity

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