1,678 reputation
1825
bio website
location
age 19
visits member for 3 years, 9 months
seen 17 hours ago

19h
comment Does the line $(2,1,1)+t(-3,1,5)$ live within the plane $31x+3y+18z=62$?
@Galc127: The formula you posted, is that for calculating the orthogonal projection? In that case, don't I have to multiply the result by $(31,3,18)$, and then calculate the magnitude of the result?
23h
accepted Does the line $(2,1,1)+t(-3,1,5)$ live within the plane $31x+3y+18z=62$?
1d
asked Does the line $(2,1,1)+t(-3,1,5)$ live within the plane $31x+3y+18z=62$?
1d
accepted The normal equation of the plane that contains the line $(1,1,1) + t(-2,0,3)$
1d
accepted Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane?
1d
comment Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane?
Would the same apply parallel vectors? I mean, if I find a vector that is parallel to that line's direction vector, would it also be parallel to the plane?
1d
asked Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane?
1d
asked The normal equation of the plane that contains the line $(1,1,1) + t(-2,0,3)$
Oct
14
awarded  Popular Question
Oct
3
accepted Deriving $\frac{8}{\sqrt{x-2}}$
Oct
2
awarded  Notable Question
Oct
2
asked Deriving $\frac{8}{\sqrt{x-2}}$
Oct
1
comment Why $(-1 \cdot h) = -1$ when $h$ approaches $0$?
By the way, if $h \not = 0$, how come the denominator is $(x-1)^2$? I had said that "$h$ is practically $0$" but I'm not that convinced anymore.
Oct
1
accepted Why $(-1 \cdot h) = -1$ when $h$ approaches $0$?
Oct
1
asked Why $(-1 \cdot h) = -1$ when $h$ approaches $0$?
Sep
29
awarded  Popular Question
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@AndréNicolas I see now. Thanks. By the way, you picked $\frac{1}{4}$ because it made arithmetic easier - but how can I determine the "largest acceptable value" that can replace $\frac{1}{4}$? (out of curiosity)
Sep
20
accepted Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@EWHLee I see. I wonder if $\frac{1}{4x-9}$ should have been $\frac{1}{|4x-9|}$ instead... Wait, no, that doesn't work either. Goddammit. What should I have done instead?
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@AndréNicolas Ahhhh yes, that's right. By the way, isn't $|(x-3)| < \delta < \frac{9}{4}$ enough to keep $|4x-9|$ away from $0$ since $\frac{9}{4}$ is the solution to $4x-9$?