Solving a Vector Problem: Finding the Midpoint of V (-3,-2,-2) and (8,6,9)

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Homework Statement


Consider the vector V between (-3,-2,-2) and (8,6,9).
What is vector V ?


Isn't this question asking just for the midpoint in that case shouldn't it be <5/2,2,7/2>. Webwork my online homework submitter tells me I'm wrong.
 
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Most likely it is asking for the vector made by the two points.
 
The MIDPOINT between two vectors A and B is (A+B)/2. I think the 'between' in this problem means the vector that joins the two points A and B. That would be B-A. Your confusion about the word 'between' is understandable.
 
Dick said:
The MIDPOINT between two vectors A and B is (A+B)/2. I think the 'between' in this problem means the vector that joins the two points A and B. That would be B-A. Your confusion about the word 'between' is understandable.

Yes, you are right thank you.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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