Equation Of A Plane Containing A Straight Line

Introduction

In the realm of Analytic Geometry, understanding the equation of a plane containing a straight line is crucial for various applications in mathematics and physics. A straight line can be defined by a system of equations, and in this article, we will explore how to find the equation of a plane that contains this line. We will use the points $(1, 2, 3)$ and $(-3, -2, -1)$ to define the straight line and then derive the equation of the plane containing this line.

Vector Representation of the Straight Line

A straight line can be represented by a vector equation, which is given by:

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

where $\vec{r}$ is the position vector of any point on the line, $\vec{r_0}$ is the position vector of a fixed point on the line, $\vec{v}$ is the direction vector of the line, and $t$ is a parameter.

In our case, the fixed point is $(1, 2, 3)$, and the direction vector can be found by subtracting the coordinates of the two given points:

$$\vec{v} = (-3 - 1, -2 - 2, -1 - 3) = (-4, -4, -4)$$

The direction vector is $(-4, -4, -4)$, which can be simplified to $(-1, -1, -1)$ by dividing all its components by $-4$. Therefore, the vector equation of the straight line is:

$$\vec{r} = (1, 2, 3) + t(-1, -1, -1)$$

System of Equations of the Straight Line

We have already derived the system of equations of the straight line as:

$$x = 1 - t,\, y = 2 - t, \, z = 3 - t$$

These equations represent the straight line in Cartesian coordinates.

Equation of a Plane Containing the Straight Line

To find the equation of a plane containing the straight line, we need to find two vectors that lie on the plane. One of these vectors is the direction vector of the straight line, which is $(-1, -1, -1)$. To find the second vector, we can use the fact that the plane contains the straight line. We can choose any point on the straight line, say $(1, 2, 3)$, and find another point on the plane. Let's call this point $(x, y, z)$. Then, the vector from $(1, 2, 3)$ to $(x, y, z)$ lies on the plane.

The vector from $(1, 2, 3)$ to $(x, y, z)$ is:

$$\vec{v_2} = (x - 1, y - 2, z - 3)$$

Since this vector lies on the plane, it is perpendicular to the normal vector of the plane. The normal vector of the plane is the cross product of the two vectors that lie on the plane. Therefore, we can find the normal vector of the plane by taking the cross product of the direction vector of the straight line and the vector from $(, 2, 3)$ to $(x, y, z)$:

$$\vec{n} = (-1, -1, -1) \times (x - 1, y - 2, z - 3)$$

Expanding the cross product, we get:

$$\vec{n} = ((y - 2)(-1) - (z - 3)(-1), (z - 3)(-1) - (x - 1)(-1), (x - 1)(-1) - (y - 2)(-1))$$

Simplifying the expression, we get:

$$\vec{n} = (-y + 2 + z - 3, -z + 3 + x - 1, -x + 1 + y - 2)$$

$$\vec{n} = (-y + z - 1, x - z + 2, -x + y - 1)$$

The normal vector of the plane is $(-y + z - 1, x - z + 2, -x + y - 1)$.

Equation of the Plane

The equation of the plane containing the straight line is given by:

$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

where $(x_0, y_0, z_0)$ is a point on the plane, and $(a, b, c)$ is the normal vector of the plane.

Substituting the values of the normal vector and the point $(1, 2, 3)$, we get:

$$(-y + z - 1)(x - 1) + (x - z + 2)(y - 2) + (-x + y - 1)(z - 3) = 0$$

Expanding the equation, we get:

$$-xy + xz - x + yz - z - y + 2y - 4 + xz - 2x + 2y - 4 - xz + y - z + 3 = 0$$

Simplifying the equation, we get:

$$-xy + yz - z - y + 2y - 4 - 2x + 2y - 4 + y - z + 3 = 0$$

$$-xy + yz - 2x + 4y - 5 = 0$$

The equation of the plane containing the straight line is $-xy + yz - 2x + 4y - 5 = 0$.

Conclusion

In this article, we have derived the equation of a plane containing a straight line. We have used the points $(1, 2, 3)$ and $(-3, -2, -1)$ to define the straight line and then found the equation of the plane containing this line. The equation of the plane is $-xy + yz - 2x + 4y - 5 = 0$. This equation represents the plane in Cartesian coordinates.

References

• [1] Anton, H. (2018). Calculus: Early Transcendentals. John Wiley & Sons.
• [2] Edwards, C. H. (2019). Differential Equations: Theory and Applications. John Wiley & Sons.
• [3] Strang, G. (2019). Linear Algebra and Its Applications. Cengage Learning.

Glossary

• **Analytic Geometry A branch of mathematics that deals with the study of geometric shapes using algebraic and analytical methods.
• Cartesian Coordinates: A system of coordinates that uses the x, y, and z axes to locate points in a three-dimensional space.
• Direction Vector: A vector that points in the direction of a line or a plane.
• Normal Vector: A vector that is perpendicular to a plane or a surface.
• Plane: A flat surface that extends infinitely in all directions.
• Straight Line: A line that extends infinitely in one direction and has no curvature.
  Equation of a Plane Containing a Straight Line: Q&A =====================================================

Q: What is the equation of a plane containing a straight line?

A: The equation of a plane containing a straight line is given by:

$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

where $(x_0, y_0, z_0)$ is a point on the plane, and $(a, b, c)$ is the normal vector of the plane.

Q: How do I find the equation of a plane containing a straight line?

A: To find the equation of a plane containing a straight line, you need to find two vectors that lie on the plane. One of these vectors is the direction vector of the straight line, and the other vector can be found by choosing any point on the straight line and finding another point on the plane.

Q: What is the normal vector of a plane?

A: The normal vector of a plane is a vector that is perpendicular to the plane. It can be found by taking the cross product of two vectors that lie on the plane.

Q: How do I find the normal vector of a plane?

A: To find the normal vector of a plane, you need to find two vectors that lie on the plane. One of these vectors is the direction vector of the straight line, and the other vector can be found by choosing any point on the straight line and finding another point on the plane. Then, take the cross product of these two vectors to find the normal vector.

Q: What is the equation of the plane containing the straight line defined by the points (1, 2, 3) and (-3, -2, -1)?

A: The equation of the plane containing the straight line defined by the points (1, 2, 3) and (-3, -2, -1) is:

$$-xy + yz - 2x + 4y - 5 = 0$$

Q: How do I use the equation of a plane to find the distance from a point to the plane?

A: To find the distance from a point to a plane, you need to use the equation of the plane and the coordinates of the point. The distance from the point to the plane is given by:

$$d = \frac{|a(x - x_0) + b(y - y_0) + c(z - z_0)|}{\sqrt{a^2 + b^2 + c^2}}$$

where $(x, y, z)$ is the point, and $(x_0, y_0, z_0)$ is a point on the plane.

Q: What is the significance of the equation of a plane in real-world applications?

A: The equation of a plane has many real-world applications, such as:

• Architecture: The equation of a plane is used to design buildings and other structures.
• Engineering: The equation of a plane is used to design bridges, roads, and other infrastructure.
• Computer Graphics: The equation of a plane is used to create 3D models and animations.
• Physics: The equation of a plane is used to the motion of objects in space.

Q: Can you provide more examples of the equation of a plane?

A: Yes, here are a few more examples of the equation of a plane:

• Plane containing the straight line defined by the points (2, 3, 4) and (5, 6, 7):

$$-xy + yz - 2x + 4y - 5 = 0$$

• Plane containing the straight line defined by the points (1, 2, 3) and (4, 5, 6):

$$-xy + yz - 2x + 4y - 5 = 0$$

• Plane containing the straight line defined by the points (3, 4, 5) and (6, 7, 8):

$$-xy + yz - 2x + 4y - 5 = 0$$

Conclusion

In this article, we have provided a comprehensive overview of the equation of a plane containing a straight line. We have discussed the equation of a plane, how to find the equation of a plane, and the significance of the equation of a plane in real-world applications. We have also provided examples of the equation of a plane and answered frequently asked questions.