Problem 53
Question
Show that the vector \(\mathbf{n}=a \mathbf{i}+b \mathbf{j}\) is perpendicular to the line with equation \(a x+b y=c .\) Hint: Let \(P_{1}\left(x_{1}, y_{1}\right)\) and \(P_{2}\left(x_{2}, y_{2}\right)\) be two points on the line and show that \(\mathbf{n} \cdot \overrightarrow{P_{1} P_{2}}=0\).
Step-by-Step Solution
Verified Answer
Vector \(\mathbf{n}\) is perpendicular to the line because the dot product \(\mathbf{n} \cdot \overrightarrow{P_{1}P_{2}} = 0\).
1Step 1: Understand the Problem
We need to show that a given vector \(\mathbf{n}=a \mathbf{i}+b \mathbf{j}\) is perpendicular to the line described by the equation \(ax + by = c\). A vector is perpendicular to a line if the dot product of the vector with the direction vector of the line is zero.
2Step 2: Identify Points on the Line
Let \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\) be two arbitrary points on the line. These points satisfy the line equation, so \( a x_1 + b y_1 = c \) and \( a x_2 + b y_2 = c \).
3Step 3: Find the Direction Vector of the Line
The direction vector \(\overrightarrow{P_{1}P_{2}}\) from \(P_1\) to \(P_2\) can be represented as \(\langle x_2 - x_1, y_2 - y_1 \rangle\).
4Step 4: Calculate the Dot Product
Now compute the dot product of \(\mathbf{n}\) and the direction vector: \(\mathbf{n} \cdot \overrightarrow{P_{1}P_{2}} = (a \mathbf{i} + b \mathbf{j}) \cdot \langle x_2 - x_1, y_2 - y_1 \rangle = a(x_2 - x_1) + b(y_2 - y_1)\).
5Step 5: Simplify the Expression
Since \(x_1\), \(y_1\), \(x_2\), and \(y_2\) lie on the line, we know \(a x_1 + b y_1 = c\) and \(a x_2 + b y_2 = c\). Thus, \(a(x_2 - x_1) + b(y_2 - y_1) = (a x_2 + b y_2) - (a x_1 + b y_1) = c - c = 0\).
6Step 6: Conclusion - Perpendicularity
We have shown that \(\mathbf{n} \cdot \overrightarrow{P_{1}P_{2}} = 0\), which confirms that the vector \(\mathbf{n}\) is indeed perpendicular to every segment defined by points on the line \(ax + by = c\).
Key Concepts
Dot ProductLine EquationsVector MathematicsAnalytic Geometry
Dot Product
The dot product is a fundamental operation in vector mathematics, providing insight into the angle between two vectors. In simpler terms, the dot product measures how much one vector goes in the direction of another.
An essential property of the dot product is related to perpendicular vectors. If two vectors are perpendicular (or orthogonal), their dot product will be zero.
This happens because the angle \( \theta \) between them is 90 degrees, and the cosine of 90 degrees is zero. Hence,
An essential property of the dot product is related to perpendicular vectors. If two vectors are perpendicular (or orthogonal), their dot product will be zero.
This happens because the angle \( \theta \) between them is 90 degrees, and the cosine of 90 degrees is zero. Hence,
- Dot product formula: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \) (for 3D vectors)
- Two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular if \( \mathbf{a} \cdot \mathbf{b} = 0 \)
Line Equations
Understanding line equations is crucial for solving problems involving vectors and geometry. A line equation in the plane can often be represented in a standard form like \( ax + by = c \).
This linear equation shows a straight line when plotted on an x-y graph, where:
This is proven by showing that the dot product of the vector \( \mathbf{n} \) with any vector parallel to the line results in zero, establishing the perpendicular nature.
This linear equation shows a straight line when plotted on an x-y graph, where:
- \(a\) and \(b\) are coefficients that determine the direction or slope of the line.
- \(c\) is the constant that shifts the line to its correct position on the graph.
This is proven by showing that the dot product of the vector \( \mathbf{n} \) with any vector parallel to the line results in zero, establishing the perpendicular nature.
Vector Mathematics
Vector mathematics is a straightforward yet incredibly powerful tool in understanding geometry and physics concepts. Vectors represent quantities that have both magnitude and direction.
In 2D space, a vector \( \mathbf{v} \) can be described with its components like \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} \).
Vectors can perform operations such as addition, subtraction, and multiplication, including the dot product.
For instance, the stark lines through a point can be described using vectors.
This makes finding relationships, such as perpendicularity, between vectors and lines easier to visualize and compute.
In 2D space, a vector \( \mathbf{v} \) can be described with its components like \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} \).
Vectors can perform operations such as addition, subtraction, and multiplication, including the dot product.
- Addition/Subtraction: Combining vectors by adding or subtracting corresponding components.
- Scalar Multiplication: Changing a vector's magnitude without affecting its direction.
For instance, the stark lines through a point can be described using vectors.
This makes finding relationships, such as perpendicularity, between vectors and lines easier to visualize and compute.
Analytic Geometry
Analytic geometry combines the use of algebra and coordinate systems to solve geometric problems efficiently.It simplifies the process of analyzing and understanding the shapes, sizes, and positions of figures using coordinates.
For example, to determine if a vector is perpendicular to a line, one combines understanding of angle, dot product, and the line equation in a simple plane.
By utilizing coordinate points, one can easily establish the vectors' directions and positions, further aiding the visualization and comprehension of their relationships.
For example, to determine if a vector is perpendicular to a line, one combines understanding of angle, dot product, and the line equation in a simple plane.
By utilizing coordinate points, one can easily establish the vectors' directions and positions, further aiding the visualization and comprehension of their relationships.
- Coordinate Systems: Provides a basis for measuring positions in space.
- Geometry and Algebra: Bridges the gap between abstract algebraic formulas and tangible geometric interpretations.
Other exercises in this chapter
Problem 52
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