Question

let the direction vector of the line be u = <p, q, r>. to find u and a second point on the line, compare the equation of the line $\frac{x}{2}=\frac{y - 4}{-1}=z$ with the standard form of the equation $\frac{x - x_1}{p}=\frac{y - y_1}{q}=\frac{z - z_1}{r}$. u = <, -1, 1> (x1, y1, z1) = (, 4, )

Explanation:

Step1: Recall the standard form of the line equation

The standard form of a line in 3 - D is $\frac{x - x_1}{p}=\frac{y - y_1}{q}=\frac{z - z_1}{r}$, where $(x_1,y_1,z_1)$ is a point on the line and $\vec{u}=(p,q,r)$ is the direction - vector of the line.

Step2: Compare the given equation $\frac{x}{2}=\frac{y - 4}{-1}=z$ with the standard form

We can rewrite the given equation as $\frac{x-0}{2}=\frac{y - 4}{-1}=\frac{z - 0}{1}$.

Step3: Determine the direction - vector and a point on the line

For the direction - vector $\vec{u}=(p,q,r)$, by comparing with the standard form, we have $p = 2$, $q=-1$, $r = 1$. So $\vec{u}=(2,-1,1)$. And a point on the line $(x_1,y_1,z_1)=(0,4,0)$.

Answer:

$u=(2,-1,1)$; $(x_1,y_1,z_1)=(0,4,0)$