In Exercises 72-78, we consider the equations of a line in...

In Exercises $72-78,$ we consider the equations of a line in symmetric form, when $a \neq 0, b \neq 0, c \neq 0$. $$ \frac{x-x_{0}}{a}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c} $$ Let $\mathcal{L}$ be the line through $P_{0}=\left(x_{0}, y_{0}, z_{0}\right)$ with direction vector $\mathbf{v}=$ $\langle a, b, c\rangle .$ Show that $\mathcal{L}$ is defined by the symmetric equations (10). Hint: Use the vector parametrization to show that every point on $\mathcal{L}$ satisfies (10).

In Exercises $72-78,$ we consider the equations of a line in symmetric form, when $a \neq 0, b \neq 0, c \neq 0$.
$$
\frac{x-x_{0}}{a}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c}
$$
Let $\mathcal{L}$ be the line through $P_{0}=\left(x_{0}, y_{0}, z_{0}\right)$ with direction vector $\mathbf{v}=$ $\langle a, b, c\rangle .$ Show that $\mathcal{L}$ is defined by the symmetric equations (10). Hint: Use the vector parametrization to show that every point on $\mathcal{L}$ satisfies (10).

Key Concepts

Vector Parametrization of a Line

This concept involves representing a line in three-dimensional space by using a point through which the line passes and a direction vector. The standard form is r(t) = r? + t·v, where r? is a position vector of a given point on the line and v is the direction vector. This representation is fundamental in understanding the geometry of lines and in analyzing the relationship between points on the line as the parameter t varies.

Direction Vector

The direction vector defines the orientation and the direction along which the line extends in space. Its components indicate how much the x, y, and z coordinates change as one moves along the line. This vector is crucial for forming the symmetric equations because its components directly appear in the denominators that relate the coordinate differences to each other.

Symmetric Equations of a Line

Derived from the vector parametrization of a line, the symmetric equations eliminate the parameter to relate the coordinates directly. The form (x ? x?)/a = (y ? y?)/b = (z ? z?)/c captures the invariant ratios dictated by the direction vector components and provides an alternative and compact representation of the line.

Elimination of the Parameter

This process involves removing the parameter from the parametric equations of a line to obtain a direct relationship among the x, y, and z coordinates. By expressing the parameter t in terms of each coordinate and equating the results, one derives the symmetric form of the line's equation, which is a key technique in analytic geometry.

Transcript

00:01 Now we are given three real numbers abc and none of them is equal to zero.

00:07 And we express some curve by equation here, by an equation here.

00:15 And let l be the straight line passing through p naught and passing through x naught, y naught, z naught with the direction vector abc.

00:26 We want to show the curve c is actually equal to the straight line l.

00:35 Okay, we know the equation of l can be written as x naught, y naught, z naught plus t, abc for any t real.

00:53 Okay, first notice when t is equal to zero, we have the fixed point p.

00:59 And it's easy to see the p naught just satisfies this equation of c.

01:09 That means p naught is contained in c.

01:12 Okay, now for any t not zero, we want to show this point is also a point in c.

01:32 Okay, then what we need to do is to plug in all the expression into c to check whether they satisfy the equation.

01:42 Okay, now our x is equal to x naught plus ta.

01:47 Y is equal to y naught plus b.

01:52 C is equal to c naught plus tc.

01:56 So x minus x naught is equal to ta.

02:00 Y minus y naught is equal to tb.

02:09 Now, if we divide a here, b here, and c here, we just get three t.

02:19 And it is equivalent to c.

02:24 Those two points, those three equations, those three relationships just satisfy our equation for the curve c or equivalently, let's say, defined as pt...

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