Find the equation of a line through the point (-2, 1, 3) and parallel to the line: \[ \frac{x - 2}{4} = \frac{y + 3}{-3}, \quad z = -2 \]
\( \frac{x+2}{4} = \frac{y-1}{-3}, z = 3 \)
Understanding the Problem: Finding the Equation of a Line
The question asks us to find the equation of a straight line in three-dimensional space. We are given two crucial pieces of information about this line:
- It must pass through a specific point: (-2, 1, 3).
- It must be parallel to another given line.
To find the equation of a line in 3D, we typically need two things: a point that the line passes through and a direction vector that indicates the line's orientation in space. We already have a point (-2, 1, 3). The key is to use the information that the line is parallel to the given line to find its direction vector.
Identifying the Direction Vector of the Given Line
The given line is described by the equations:
\[ \frac{x - 2}{4} = \frac{y + 3}{-3}, \quad z = -2 \]This form tells us about the line's direction. The equation \(\frac{x - 2}{4} = \frac{y + 3}{-3}\) is part of the symmetric form \(\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\), where \(\langle a, b, c \rangle\) is the direction vector. From the given part \(\frac{x - 2}{4} = \frac{y + 3}{-3}\), we can see that the components of the direction vector in the x and y directions are 4 and -3, respectively.
The second part, \(z = -2\), is a constraint on the z-coordinate. It means that the z-coordinate of any point on this line is always -2. This implies that the line is parallel to the xy-plane. If the z-coordinate never changes, the component of the direction vector in the z-direction must be 0.
So, the direction vector of the given line is \(\vec{v}_{given} = \langle 4, -3, 0 \rangle\).
Using Parallelism to Find the Direction Vector of Our Line
When two lines are parallel in 3D space, they share the same direction vector (or a scalar multiple of it). Since the line we are looking for is parallel to the given line, it will have the same direction vector.
Therefore, the direction vector for the line we need to find is \(\vec{v} = \langle 4, -3, 0 \rangle\).
Constructing the Equation of the Line
We have the point the line passes through, \(P_0(-2, 1, 3)\), and the direction vector, \(\vec{v} = \langle 4, -3, 0 \rangle\). We can write the equation of the line in symmetric form:
\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \]Substituting the point \((x_0, y_0, z_0) = (-2, 1, 3)\) and the direction vector \(\langle a, b, c \rangle = \langle 4, -3, 0 \rangle\):
\[ \frac{x - (-2)}{4} = \frac{y - 1}{-3} = \frac{z - 3}{0} \]This becomes:
\[ \frac{x + 2}{4} = \frac{y - 1}{-3} = \frac{z - 3}{0} \]However, division by zero is undefined. The form \(\frac{z - z_0}{0}\) in the symmetric equation specifically means that the z-component of the direction vector is 0, which implies that the z-coordinate of any point on the line is constant and equal to \(z_0\). In this case, \(z_0 = 3\).
So, the equation of the line is represented by the relation between x and y coordinates and a separate equation for the z coordinate:
\[ \frac{x + 2}{4} = \frac{y - 1}{-3}, \quad z = 3 \]Comparing with the Options
Let's compare our derived equation with the given options:
- Option 1: \( \frac{x+2}{4} = \frac{y-1}{-3}, z = 3 \)
- Option 2: \( \frac{x+2}{4} = \frac{y-1}{-3}, z = -3 \)
- Option 3: \( \frac{x+2}{4} = \frac{y-1}{3}, \frac{z-3}{1} \)
- Option 4: \( \frac{x+2}{4} = \frac{y-1}{3} = \frac{z+3}{1} \)
Our derived equation \( \frac{x+2}{4} = \frac{y-1}{-3}, z = 3 \) exactly matches Option 1.
Conclusion
The equation of the line passing through (-2, 1, 3) and parallel to \( \frac{x - 2}{4} = \frac{y + 3}{-3}, \quad z = -2 \) is \( \frac{x+2}{4} = \frac{y-1}{-3}, z = 3 \).
Revision Table: Key Concepts for Line Equations
| Concept | Description | Formulas |
|---|---|---|
| Direction Vector | A vector indicating the line's direction. Parallel lines have proportional direction vectors. | \(\vec{v} = \langle a, b, c \rangle\) |
| Point on Line | Any single point that lies on the line. | \(P_0(x_0, y_0, z_0)\) |
| Symmetric Form | Relates coordinates x, y, z using direction ratios, provided none are zero. | \(\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\) (if \(a,b,c \ne 0\)) |
| Case: Direction Ratio is Zero | If a direction ratio (e.g., c) is zero, that coordinate (e.g., z) is constant. | e.g., if \(c=0\), then \(z=z_0\). Equation becomes \(\frac{x - x_0}{a} = \frac{y - y_0}{b}, z = z_0\) |
| Vector Form | Uses a position vector to a point on the line and the direction vector. | \(\vec{r} = \vec{r_0} + t\vec{v}\) |
| Parametric Form | Expresses x, y, z coordinates in terms of a parameter \(t\). | \(x = x_0 + at\), \(y = y_0 + bt\), \(z = z_0 + ct\) |
Additional Information: Types of Line Equations in 3D
Understanding the different ways to represent a line in 3D space is essential for solving problems like this one. The main forms are vector form, parametric form, and symmetric form.
- Vector Form: A line passing through point \(A\) with position vector \(\vec{r_0}\) and parallel to vector \(\vec{v}\) has the equation \(\vec{r} = \vec{r_0} + t\vec{v}\), where \(\vec{r}\) is the position vector of any point on the line and \(t\) is a scalar parameter.
- Parametric Form: If \(\vec{r} = \langle x, y, z \rangle\), \(\vec{r_0} = \langle x_0, y_0, z_0 \rangle\), and \(\vec{v} = \langle a, b, c \rangle\), the vector equation expands to \(\langle x, y, z \rangle = \langle x_0, y_0, z_0 \rangle + t\langle a, b, c \rangle\). This gives the parametric equations: \[ x = x_0 + at \] \[ y = y_0 + bt \] \[ z = z_0 + ct \] These equations express the coordinates of any point on the line in terms of the parameter \(t\).
- Symmetric Form: If \(a, b, c\) are all non-zero, we can solve each parametric equation for \(t\) and set them equal: \[ t = \frac{x - x_0}{a} \] \[ t = \frac{y - y_0}{b} \] \[ t = \frac{z - z_0}{c} \] This leads to the symmetric form: \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \] If one or more of \(a, b, c\) are zero, the symmetric form is modified as shown in the solution above (e.g., if \(c=0\), the equation is \(\frac{x - x_0}{a} = \frac{y - y_0}{b}\) and \(z = z_0\)). This indicates that the line lies in a plane parallel to one of the coordinate planes.
In this problem, the direction vector \(\langle 4, -3, 0 \rangle\) has a zero component, which is why the resulting equation correctly uses the modified symmetric form combined with a constant z-value.
Similar Questions
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If the distance of the point (4,6,8) from the plane \( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \) is 1, then \( a \) is:
If the equation of a line \( PQ \) is:
\[ \frac{x+1}{2} = \frac{2-y}{5} = \frac{z+6}{7} \]
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Important Questions from Three-Dimensional Geometry
There are two bags. Bag-1 contains 4 white and 6 black balls and Bag-2 contains 5 white and 5 black balls.
A die is rolled. If it shows a number divisible by 3, a ball is drawn from Bag-1; otherwise, a ball is drawn from Bag-2.
If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is:
The unit vector perpendicular to each of the vectors $ \vec{a} + \vec{b}$ and $ \vec{a} - \vec{b}$, where, $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is :
The direction cosines of the line which is perpendicular to the lines with direction ratios (1, -2, -2) and (0, 2, 1) are:
The tangent to the circle centered at (0,0) with radius 1 at point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) is given by:
If the distance of the point (4,6,8) from the plane \( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \) is 1, then \( a \) is:
