Question

Find the symmetric equations of the line through \((-5,7,-2)\) and perpendicular to both \(\langle 2,1,-3\rangle\) and \(\langle 5,4,-1\rangle\).

Short answer

\(\frac{x + 5}{8} = \frac{y - 7}{-13} = \frac{z + 2}{3}\).

Step-by-step solution

Find the Direction Vector using Cross Product

To find a line perpendicular to both given vectors, we use the cross product of the vectors \( \langle 2, 1, -3 \rangle \) and \( \langle 5, 4, -1 \rangle \). The cross product is calculated as follows:\[\mathbf{v} = \langle 2, 1, -3 \rangle \times \langle 5, 4, -1 \rangle = \langle (-1)(4) - (-3)(4), (-3)(5) - (2)(-1), (2)(4) - (1)(5) \rangle\]So we compute:- First component: \(-4 + 12 = 8\)- Second component: \(-15 + 2 = -13\)- Third component: \(8 - 5 = 3\)Thus, the direction vector is \( \mathbf{v} = \langle 8, -13, 3 \rangle \).

Write the Symmetric Equations

We have the direction vector \( \langle 8, -13, 3 \rangle \) and a point on the line, \((-5, 7, -2)\). The symmetric equations of a line given a point \((x_0, y_0, z_0)\) and a direction vector \( \langle a, b, c \rangle \) are:\[\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\]Substituting in our values:\[\frac{x + 5}{8} = \frac{y - 7}{-13} = \frac{z + 2}{3}\]This gives us the symmetric equations of the line.

Key concepts

Understanding the Cross Product

The cross product is a critical concept in vector calculus and is mainly used when dealing with three-dimensional vectors. This operation helps us find a vector that is perpendicular (or normal) to two given vectors. When given two vectors, say \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a new vector \( \mathbf{c} = \langle c_1, c_2, c_3 \rangle \). Each component of this new vector is determined using formulas that involve the determinants:

• First component: \( c_1 = a_2b_3 - a_3b_2 \)
• Second component: \( c_2 = a_3b_1 - a_1b_3 \)
• Third component: \( c_3 = a_1b_2 - a_2b_1 \)

These calculations ensure \( \mathbf{c} \) is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). For our example, the cross product gives us the perpendicular direction vector \( \langle 8, -13, 3 \rangle \). Thus, by applying this method, we can efficiently find vectors that satisfy specific perpendicularity conditions.
It's a handy tool in finding the direction vector necessary for constructing symmetric equations of a line.

Direction Vector in Line Equations

A direction vector is a fundamental part of defining the equation of a line in vector calculus. Given a line, the direction vector indicates the direction in which the line extends. It's essentially a vector parallel to the line, and it defines how you "move" from one point on the line to another.
In our exercise, after determining the cross product of the given vectors, we acquired the direction vector \( \langle 8, -13, 3 \rangle \). This vector serves as a building block for formulating symmetric equations of the line.

• Direction vectors can help describe lines in both two-dimensional and three-dimensional spaces.
• When you have a direction vector, you can easily find other points on the line using the point-direction form.
• The components of the vector show its influences along the x, y, and z axes.

Understanding this concept and calculating a valid direction vector is vital for tasks like finding lines that are parallel, coincident, or perpendicular to other lines or planes.

Understanding Perpendicular Lines

Perpendicular lines have the property that they intersect at a right angle, forming a 90-degree intersection. In vector calculus, for two lines to be perpendicular, the dot product of their direction vectors must be zero.
Let's explore how this plays out in our context:

• Given two initial vectors, finding a third vector that is perpendicular to both often involves using the cross product, as demonstrated in this exercise.
• When vectors are perpendicular, it simplifies the systems we work with, especially when defining planes and lines.
• This is particularly crucial in 3D space, where calculating angles between vectors and the orientation of geometric objects is vital.

By using the cross product, we calculated a direction vector that was perpendicular to both given vectors, thus finding a line that was perpendicular to the plane created by those vectors. Consequently, understanding perpendicular concepts helps in efficiently describing spatial relationships and orientations.

Basics of Vector Calculus

Vector calculus is a field of mathematics that focuses on vector fields and differentiable functions over spaces. It combines aspects of calculus and vectors to handle problems involving differentiation and integration of vector quantities.
Key areas of vector calculus that are relevant to this exercise include:

• Vector Addition: Combining two vectors results in a new vector found through component-wise addition.
• Scalar Multiplication: Scaling a vector by multiplying it with a scalar, affecting its magnitude but not its direction.
• Dot and Cross Products: Important operations that can find angles between vectors (dot product) or perpendicular vectors (cross product).

In geometrical settings, like our exercise, vector calculus allows for the determination of directional and perpendicular lines. This is crucial for many applications in science and engineering, where spatial data and directional control are frequently required. Embracing vector calculus concepts is essential for mastering modern mathematical problems that involve vectors and directional fields.