Question

Find a parametrization
r(t) = h x(t), y(t), z(t)i
of the straight line passing through the origin
in 3-space whose projection on the xy-plane
is a line with slope 4, while its projection on
the yz-plane is a line with slope −3, i.e.,
∆y/∆x= 4,∆z/∆y= −3.

Answer 1

Can you come up with a vector which goes in the correct direction?

Answer 2

How is that?

Answer 3

First, it doesn't matter what the magnitude of the vector is... so we get to chose one of the components as long as the others are correctly proportional.

Answer 4

Our vector only has to have the right direction.

Answer 5

@theanonymous27 Does this make sense?

Answer 6

I understand, but how can I find that vector? what I dont get is how can I use the slopes here to find what I need

Answer 7

So we can start by letting our \(x\) component be 1:\[
\mathbf{v}
=
\begin{bmatrix}
x_1 \\
y_1 \\
z_1
\end{bmatrix}
=
\begin{bmatrix}
1 \\
y_1 \\
z_1
\end{bmatrix}
\]

Answer 8

Since \(\Delta y/\Delta x= 4\) and \(\Delta x = 1\), what must \(\Delta y\) be?

Answer 9

We use the slopes to figure out the proportions between each component, @theanonymous27

Answer 10

so <Y is 4, and then we get that Z has to be -12

Answer 11

Yes

Answer 12

so does the slope represent the component of the vector?

Answer 13

we end up with r(t) = <t, 4, -12t> ? is that right?

Answer 14

So given the direction vector \(\mathbf{v}\) and some point which the line goes though \(\mathbf{b}\) the parametrization for our line is: \[
\mathbf{r}(t) = \mathbf{v}t+\mathbf{b}
\]

Answer 15

\[
r(t)
=
\begin{bmatrix}
1 \\
4 \\
-12
\end{bmatrix}
t
+
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}
=
\begin{bmatrix}
t \\
4t \\
-12t
\end{bmatrix}
\]

Answer 16

Thank you!, One last question... Why did you pick x=1 first? what would have happened if you choose y=1 instead

Answer 17

If you pick \(y=1\) then you get \(x=1/4\) and \(z=-3\)

Answer 18

It's the same direction.

Answer 19

Thanks!