Calculus: equations of lines and planes

Question

If $$\frac{x-2}{4} = \frac{y-4}{-2} = z$$ is the symmetric equation for a line, what are the coordinates where it crosses the $yz$-plane, and what would the vector equation for this line be?

Answer 1

When the line crosses the $yz$-plane, its $x$ coordinate is $0$. $$\frac{-2}{4}=\frac{y-4}{-2}=z$$ Solve for $y, z$ then $y=5, z=-\frac{1}{2}$.

The vector form is $$\frac{x-2}{4}=\frac{y-4}{-2}=z=t$$ solve for $x, y, z$ and express it as a function of $t$. $$x=4t+2, y=-2t+4, z=t$$ Thus, $$X(t)=(2, 4, 0)+(4, -2, 1)t \quad t\in\mathbb{R}$$

Answer 2

1) Find the point where the line $\frac{x-2}{4} = \frac{y-4}{-2} = z$ crosses the yz-plane.

Firstly, the yz-plane is $ x = 0 $. This implies that if we let $x = 0 $ for the line $\frac{x-2}{4} = \frac{y-4}{-2} = z$, then we will be able to find where the line intersects the plane. So,

$$ \begin{align} x=0 & \implies \frac{-2}{4} = \frac{y-4}{-2} \\ & \implies y = 5 \\ & \implies \frac{5 - 4}{-2} = z\\ & \implies z = -\frac{1}{2} \\ & \end{align} $$

Therefore, the line intersects the yz-plane at the point $(0,5,-\frac{1}{2})$

2) Write the line $\frac{x-2}{4} = \frac{y-4}{-2} = z$ in vector form.

In order to write a line in vector form you need to use the vector equation of a line.

$$\vec{r} = \vec{r}_o + t\vec{v}$$

where, $\vec{r}_0$ is the vector in standard position whose terminal point exists on the line. And, $\vec{v}$ is a vector that exists parallel to the line. The paramater $t$ is introduced as a scalar multiple of $\vec{v}$. Note that a scalar multiple of a vector will always be parallel to its unit vector.

$$ \begin{align} \vec{r} = \vec{r}_o + t\vec{v}& \implies \langle x, y, z \rangle = \langle x_0, y_0, z_0 \rangle + t\langle a, b, c \rangle\\ & \implies \langle x, y, z \rangle = \langle x_0 + at, y_0 + bt, z_0 + ct \rangle \\ & \\ & \\ & \end{align} $$

Now we must find a vector equation of the line. To do this we must first re-parameterize the equation be setting it all equal to some variable, in this case lets call it $t$. So,

$$ \begin{align} t = \frac{x-2}{4} = \frac{y-4}{-2} = z & \implies z = t\\ & \implies x = 2 + 4t \\ & \implies y = 4 -2t\\ & \\ & \end{align} $$

Thus, your vector equation of the line is $\vec{r} = \langle 2 + 4t, 4 - 2t, t \rangle$