To derive the equation of a line in three-dimensional geometry through a given point and parallel to a given vector, we use the vector form and the parametric equations. Here's the detailed derivation:

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1. Representation of the Line

A line in three-dimensional space can be represented if:

• A point $P_0(x_0, y_0, z_0)$ lies on the line.
• A vector $\mathbf{v} = \langle a, b, c \rangle$ (direction vector) is parallel to the line.

The line can be expressed as the locus of all points $P(x, y, z)$ satisfying:

$\mathbf{r} = \mathbf{r_0} + t\mathbf{v}$

Where:

• $\mathbf{r}$ is the position vector of $P(x, y, z)$: $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$,
• $\mathbf{r_0}$ is the position vector of $P_0(x_0, y_0, z_0)$: $\mathbf{r_0} = x_0\mathbf{i} + y_0\mathbf{j} + z_0\mathbf{k}$,
• $\mathbf{v}$ is the given vector parallel to the line: $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$,
• $t$ is a scalar parameter.

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2. Vector Equation of the Line

From the above, the vector equation of the line is:

$\mathbf{r} = \mathbf{r_0} + t\mathbf{v}$

Substituting $\mathbf{r_0}$ and $\mathbf{v}$, we get:

$x\mathbf{i} + y\mathbf{j} + z\mathbf{k} = (x_0\mathbf{i} + y_0\mathbf{j} + z_0\mathbf{k}) + t(a\mathbf{i} + b\mathbf{j} + c\mathbf{k})$

Equating components:

$x = x_0 + ta, \quad y = y_0 + tb, \quad z = z_0 + tc$

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3. Parametric Form of the Line

From the vector equation, the parametric form is:

$x = x_0 + ta$ $y = y_0 + tb$ $z = z_0 + tc$

Where $t$ is the scalar parameter.

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4. Cartesian Form of the Line

To eliminate $t$ and express the equation in Cartesian form, solve for $t$ from each parametric equation:

$t = \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

Thus, the Cartesian equation of the line is:

$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

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Summary:

• Vector Form: $\mathbf{r} = \mathbf{r_0} + t\mathbf{v}$,
• Parametric Form: $x = x_0 + ta, \, y = y_0 + tb, \, z = z_0 + tc$,
• Cartesian Form: $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.

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Would you like more details or clarifications?

Related Questions:

1. How can the direction cosines of the vector $\mathbf{v}$ be used in the equation of a line?
2. Derive the distance of a point from a line in three-dimensional space.
3. How does the equation of a line change if the direction vector is perpendicular to another vector?
4. Discuss how to find the intersection of two lines in three dimensions.
5. Can a line in 3D be represented uniquely without a direction vector? Why or why not?

Tip: Always verify if the direction vector $\mathbf{v}$ is non-zero to avoid undefined behavior in equations.