How to find if a vector is parallel to another

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Okay, so I have two vectors a = <-6, 9, -3> and b = <2, -3, 1>. How can I test to see if these two vectors are parallel or not?
Thanks
 
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Vectors a and b are parallel if there exists a real number c such that a=cb. In your case it is pretty obvious.
 
What if it is not so obvious like the one I presented. Is there another approach to it (cross product of dot product ??)
Thanks
 
Think about the definition of the cross product, and see what happens. :smile: Btw, just use axb=det \left(\begin{array}{ccc}\vec{i} &amp; \vec{j} &amp; \vec{k}\\a_{1} &amp; a_{2} &amp; a_{3}\\b_{1} &amp; b_{2} &amp; b_{3}\end{array}\right) for a=cb.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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