Question

Find the symmetric equations of the line through \((2,-4,5)\) that is parallel to the plane \(3 x+y-2 z=5\) and perpendicular to the line $$ \frac{x+8}{2}=\frac{y-5}{3}=\frac{z-1}{-1} $$

Short answer

The symmetric equations are \( \frac{x-2}{5} = \frac{y+4}{-1} = \frac{z-5}{7} \).

Step-by-step solution

Find a Normal Vector to the Plane

The given plane equation is \(3x + y - 2z = 5\). The normal vector \( \mathbf{n} \) to the plane can be directly read from the coefficients of \(x\), \(y\), and \(z\). Therefore, \( \mathbf{n} = \langle 3, 1, -2 \rangle \).

Determine the Direction Vector of the Given Line

The line is given by the symmetric equations \( \frac{x+8}{2} = \frac{y-5}{3} = \frac{z-1}{-1} \). The direction vector \( \mathbf{d}_1 \) from these equations is \( \langle 2, 3, -1 \rangle \).

Find the Direction Vector of the Desired Line

The desired line should be parallel to the plane and thus perpendicular to its normal vector \( \mathbf{n} = \langle 3, 1, -2 \rangle \), and it should also be perpendicular to \( \mathbf{d}_1 = \langle 2, 3, -1 \rangle \).To find such a vector, we find the cross product: \[\mathbf{v} = \mathbf{n} \times \mathbf{d}_1 = \langle 3, 1, -2 \rangle \times \langle 2, 3, -1 \rangle\] This results in:\[ \mathbf{v} = \langle (1)(-1) - (-2)(3), (-2)(2) - (3)(-1), (3)(3) - (1)(2) \rangle = \langle -1 + 6, -4 + 3, 9 - 2 \rangle = \langle 5, -1, 7 \rangle.\]

Write the Symmetric Equations of the Line

The line passes through the point \((2, -4, 5)\) and has the direction vector \(\langle 5, -1, 7\rangle\). Using the symmetric form for a line, the equations are given by:\[\frac{x - 2}{5} = \frac{y + 4}{-1} = \frac{z - 5}{7}\]

Final Symmetric Equations

Thus, the symmetric equations of the required line are:\[\frac{x - 2}{5} = \frac{y + 4}{-1} = \frac{z - 5}{7}\] These equations describe a line that is parallel to the given plane and perpendicular to the given line.

Key concepts

Vector Calculus

Understanding vector calculus is essential when dealing with lines and planes in three-dimensional space. This branch of mathematics extends the concepts of calculus from scalar functions to vector fields. In the context of this exercise, it enables us to describe the orientation and relationship between different geometric entities.
Vectors are foundational in vector calculus, representing quantities with both magnitude and direction, such as velocity or force. They are often denoted using bold notation or angle brackets, like \(\langle a, b, c \rangle\).
In problems involving lines or planes, we utilize vectors for various calculations:

• Normal vectors to describe the orientation of planes.
• Direction vectors to indicate the direction of a line.

Vector calculus provides the tools to solve complex spatial problems through operations like addition, scalar multiplication, and the dot and cross products.

Normal Vector

A normal vector is a vital concept in linear algebra and geometry. It is perpendicular to a surface or a plane. In this exercise, the normal vector \(\mathbf{n} = \langle 3, 1, -2 \rangle\) directly arises from the plane's equation \(3x + y - 2z = 5\).
The coefficients of each variable in the plane's equation become the components of the normal vector:

• The coefficient of \(x\) is 3.
• The coefficient of \(y\) is 1.
• The coefficient of \(z\) is -2.

This normal vector is crucial in determining whether another line is parallel or perpendicular to the plane. In our problem, the desired line needed to be perpendicular to this normal vector to confirm it was parallel to the plane, given that perpendicularity implies a lack of intersection except at infinity.

Cross Product

The cross product is an operation between two vectors that yields another vector perpendicular to the plane containing them. In our exercise, the cross product helps us find a direction vector that satisfies given conditions relative to both a plane and another line.
Suppose we have vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\). The cross product \(\mathbf{u} \times \mathbf{v}\) results in:\[ \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle \]In the solution, we calculated the cross product of the normal vector \(\mathbf{n} = \langle 3, 1, -2 \rangle\) and the direction vector of the given line \(\mathbf{d}_1 = \langle 2, 3, -1 \rangle\).
The result, \(\mathbf{v} = \langle 5, -1, 7 \rangle\), becomes our direction vector for the line needed in the task; perpendicular to both \(\mathbf{n}\) and \(\mathbf{d}_1\), thus rendering it parallel to the given plane.

Direction Vector

A direction vector indicates the direction of a line in three-dimensional space. In the symmetric equation of a line, it shows how the points on the line displace relative to the coordinates.
The direction vector is denoted by the variables in the denominators of the symmetric equations for a line. In our exercise, we ultimately sought a line with a direction vector of \(\langle 5, -1, 7 \rangle\).
This vector was obtained using the cross product, as discussed earlier. Its components indicate how units are moved along the x, y, and z axes respectively:

• 5 units along the x-axis.
• -1 unit along the y-axis.
• 7 units along the z-axis.

This direction vector enabled us to write the symmetric equations \(\frac{x - 2}{5} = \frac{y + 4}{-1} = \frac{z - 5}{7}\), connecting it to the point \(2, -4, 5\) through which the line passes.