# First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

#### leehuan

##### Well-Known Member

Do 1911 then, don't do 1917 if you don't like computing lol.
Lol I'll analyse it economically.

I don't like lots of things. But give me a push that I enjoy and I'll probably give it a try.

Here: Well yeah, I'll probably go for 1911 at the end of the day because ENGG1811 sounds wasteful for me.
Basically, the opportunity cost of doing COMP1917 is my WAM.

I find courses far more interesting when I have someone to do it with though. For my courses this sem I basically had to wait till week 5 before I had some proper friends (except maybe Kaido for MATH1151 calculus).

If I have someone I know doing COMP1917 I'll feel more motivated to do it. The opportunity cost then becomes be a loner (in my own eyes) as well as the WAM. And the decision making process gets tougher.

(Still, 80% of my brain is tending towards COMP1911.)

The question says:

$\bg_white \text{Find the equation of the line \mathit{L} passing through P(3,-1,2) and Q(1,1,-3).}$

<x,y,z> = <2,-2,5> + <1,1,-3>t (my answer)

the same as:

<x,y,z> = <1,1,-3> + <-2,2,-5>t (textbook answer)
$\bg_white I know that the answer's already been addressed, however in general, the line passing through the point P and Q with coordinate vectors \textbf{p} and \textbf{q} is given$

$\bg_white \textbf{x}=\textbf{p}+t(\textbf{q}-\textbf{p})$

$\bg_white So since your question we have \textbf{q}-\textbf{p}=\begin{pmatrix}-2\\2\\-5\end{pmatrix} we give our answer as the textbook shows it.$

For some real number t of course.

#### turntaker

##### Well-Known Member

thanks leehuan I already understood my mistake

#### SpiralFlex

##### Well-Known Member

You're missing out on a wonderful way to see things leehuan is 1917

The courses is probably one of the most useful first year courses in UNSW.

#### leehuan

##### Well-Known Member

You're missing out on a wonderful way to see things leehuan is 1917

The courses is probably one of the most useful first year courses in UNSW.
It really is not luring if the course poses more risk to my marks than benefit to my skills set. If I end up doing 1917 some other things would've had to lure me into it.

#### SpiralFlex

##### Well-Known Member

It really is not luring if the course poses more risk to my marks than benefit to my skills set. If I end up doing 1917 some other things would've had to lure me into it.
Even in that case it's still better to do 1917 if you're concerned about marks. It's always better to do higher versions of any course at Uni if you're gunning for a solid mark.

#### leehuan

##### Well-Known Member

Even in that case it's still better to do 1917 if you're concerned about marks. It's always better to do higher versions of any course at Uni if you're gunning for a solid mark.
Without knowing how much of an effect scaling has I really don't feel like using that as my backup for underperforming in the more demanding course.

I can't say for sure if I'll end up putting too much effort into 1917 and neglecting my three other courses during the semester either, because I "failed to see a need to balance my study pattern" or etc.

#### SpiralFlex

##### Well-Known Member

Without knowing how much of an effect scaling has I really don't feel like using that as my backup for underperforming in the more demanding course.

I can't say for sure if I'll end up putting too much effort into 1917 and neglecting my three other courses during the semester either, because I "failed to see a need to balance my study pattern" or etc.
Have you had 1151 exam yet? How was it?

#### leehuan

##### Well-Known Member

Have you had 1151 exam yet? How was it?
Upcoming Friday. (Mentally not looking forward to it, physically just semi-putting it off for the time being to study for accounting)

#### Flop21

##### Well-Known Member

It really is not luring if the course poses more risk to my marks than benefit to my skills set. If I end up doing 1917 some other things would've had to lure me into it.
You can easily do well in comp1917 if you put the time and effort into it (and see the joy of it).

But it is a very time consuming course. You will need to make sure you're not neglecting any subjects while doing it. And of course comp1917 itself. If you don't keep up, then yeah your mark will suffer.

#### leehuan

##### Well-Known Member

You can easily do well in comp1917 if you put the time and effort into it (and see the joy of it).

But it is a very time consuming course. You will need to make sure you're not neglecting any subjects while doing it. And of course comp1917 itself. If you don't keep up, then yeah your mark will suffer.
I suppose you could say that about any course; effort pays off

I still need to look further into it. Thank goodness there's still at least one semester before I make a decision - 0% chance i'm gonna do any computing units next semester

#### turntaker

##### Well-Known Member

How do I convert these to vector parametric form?

#### InteGrand

##### Well-Known Member

How do I convert these to vector parametric form?
$\bg_white \noindent If the line is \frac{x-a}{v_1}=\frac{y-b}{v_2}=\frac{z-c}{v_3}, then the line is \begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}a\\b\\c\end{bmatrix}+ t \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}.$

$\bg_white \noindent To see this, since each of those fractions are equal, we can say they are all equal to t. So \frac{x-a}{v_1}=t\Rightarrow x = a + tv_1. Similarly, y=b+tv_2 and z=c+tv_3. So$

\bg_white \begin{align*} \begin{bmatrix}x \\y \\ z \end{bmatrix} &= \begin{bmatrix} a + tv_1 \\ b+tv_2 \\ c+tv_3 \end{bmatrix} \\ &= \begin{bmatrix} a \\b \\ c \end{bmatrix} + t \begin{bmatrix} v_1 \\v_2 \\v_3\end{bmatrix}.\end{align*}

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#### Flop21

##### Well-Known Member

How do I convert these to vector parametric form?
This may help:

https://youtu.be/h2MlchG06AQ?t=55s

-
Side note, no idea why he chooses to do live streams and uploads that vs just recording the thing in high quality.

#### turntaker

##### Well-Known Member

$\bg_white \noindent If the line is \frac{x-a}{v_1}=\frac{y-b}{v_2}=\frac{z-c}{v_3}, then the line is \begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}a\\b\\c\end{bmatrix}+ t \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}.$

$\bg_white \noindent To see this, since each of those fractions are equal, we can say they are all equal to t. So \frac{x-a}{v_1}=t\Rightarrow x = a + tv_1. Similarly, y=b+tv_2 and z=c+tv_3. So$

\bg_white \begin{align*} \begin{bmatrix}x \\y \\ z \end{bmatrix} &= \begin{bmatrix} a + tv_1 \\ b+tv_2 \\ c+tv_3 \end{bmatrix} \\ &= \begin{bmatrix} a \\b \\ c \end{bmatrix} + t \begin{bmatrix} v_1 \\v_2 \\v_3\end{bmatrix}.\end{align*}
Oh i understand now thanks Integrand

#### turntaker

##### Well-Known Member

is there an easy way to calculate the x,y,z plane intercepts of a 3D line

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#### InteGrand

##### Well-Known Member

is there an easy way to calculate the x,y,z intercepts of a 3D line
To find the intercept with one of the axes, set the other two variables to 0 and try and solve for the variable whose axis' intercept we're finding. (If you get no solution, it means the line doesn't intersect that axis. A line in 3D need not have any axial intercepts (in fact usually won't), unlike in 2D. You could also get infinitely many solutions, which would mean the line is actually one of the axes).

E.g. Consider the line (x,y,z) = (1,2,3) + t(4,5,6). To try finding the x-intercept if there is one, we want

(x,0,0) = (1,2,3) + t(4,5,6). So we have three equations in two unknowns (so you can see why usually we'll have no solutions):

x = 1 + 4t
0 = 2 + 5t
0 = 3 + 6t.

In this case, this system is inconsistent (see last two equations), so the line has no x-intercept. If the system was consistent though, we'd be able to solve for x and thus find our answer of (x,0,0) for the x-intercept.

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#### turntaker

##### Well-Known Member

I tried applying your given method for the line, but the textbook has a value for he intercepts. I don't think I understood the question correctly.

$\bg_white \text{Find the points where L cuts each co-ordinate plane}\linebreak L: \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3\end{bmatrix} + \begin{bmatrix}2 \\ 3 \\ 5\end{bmatrix}t$

$\bg_white (0,\frac{1}{2},\frac{-11}{2}),(\frac{-1}{3},0,\frac{-19}{3}), (\frac{11}{5},\frac{19}{5},0)$

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#### InteGrand

##### Well-Known Member

I tried applying your given method for the line, but the textbook has a value for he intercepts. I don't think I understood the question correctly.

$\bg_white \text{Find the points where L cuts each co-ordinate plane}\linebreak L: \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3\end{bmatrix} + \begin{bmatrix}2 \\ 3 \\ 5\end{bmatrix}t$

$\bg_white (0,\frac{1}{2},\frac{-11}{2}),(\frac{-1}{3},0,\frac{-19}{3}), (\frac{11}{5},\frac{19}{5},0)$
This question is asking for when it intersects the coordinate planes, rather than coordinate axes, which is what you originally asked for.

But it's a similar idea. To find the place where it cuts the xy-plane for instance, we'd set z = 0 (because the xy-plane is the place where z = 0), and then try and solve for x and y.

Any line in 3-space will cut at least one of the coordinate planes.

#### InteGrand

##### Well-Known Member

I tried applying your given method for the line, but the textbook has a value for he intercepts. I don't think I understood the question correctly.

$\bg_white \text{Find the points where L cuts each co-ordinate plane}\linebreak L: \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3\end{bmatrix} + \begin{bmatrix}2 \\ 3 \\ 5\end{bmatrix}t$

$\bg_white (0,\frac{1}{2},\frac{-11}{2}),(\frac{-1}{3},0,\frac{-19}{3}), (\frac{11}{5},\frac{19}{5},0)$
$\bg_white \noindent So for example, let's set up how to find the intersection with the xz-plane. This is where y=0, so we're looking for a point on the line with y=0, i.e. we want \begin{bmatrix} 1 \\ 2 \\ -3\end{bmatrix} + t\begin{bmatrix}2 \\ 3 \\ 5\end{bmatrix}= \begin{bmatrix}x \\ 0 \\ z\end{bmatrix} .$
$\bg_white \noindent This gives us three equations in three unknowns (namely t,x and z) to solve for. If we can solve them, then we can find the solutions for x and z and the intersection with the xz-plane will be \left(x,0,z\right). It may be the case that some of the coordinate planes don't have any intersections with a given line (which would be the case when your three equations in the three unknowns had no solutions), but at least one of them must have a solution. (If you had infinitely many solutions for one of them, it means the line actually lies in that plane.)$