Find Point on 3D Line Through Given Coordinates

Bucky
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As part of a larger question, i need to find a point the following line goes through:


r = -i + 2j + k + t(i-2k)

I have rearranged it into an alternative form (as per my notes) as

r = (t-1) i + 2j + (1-2t) k

i am supposed to be able to find a point the line passes through from this but i just can't see it, can someone help me out?
 
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What happens when t=0 or t=1?
 
see that was what i was trying to think in terms of...but i couldn't see how there were two "limiting vectors" that defined the line.

if t=0 you get

-i+2j+k

and if t=1

2j-k


i can't help but think that these are significant somehow...
 
Bucky said:
see that was what i was trying to think in terms of...but i couldn't see how there were two "limiting vectors" that defined the line.

if t=0 you get

-i+2j+k

and if t=1

2j-k


i can't help but think that these are significant somehow...
Those are the position vectors of the two points, for t=0 and t=1, through which the line described by r passes.
 
ah i think i get it. thank you very much for the help.
 
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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