Problem 20
Question
For what numbers \(c\) are \(2 c \mathbf{i}-8 \mathbf{j}\) and \(3 \mathbf{i}+c \mathbf{j}\) orthogonal?
Step-by-Step Solution
Verified Answer
The number \( c \) is 0.
1Step 1: Understanding Orthogonality Definition
Two vectors are orthogonal if their dot product equals zero. Given vectors are \( \mathbf{v_1} = 2c \mathbf{i} - 8 \mathbf{j} \) and \( \mathbf{v_2} = 3 \mathbf{i} + c \mathbf{j} \). We need to find \( c \) such that \( \mathbf{v_1} \cdot \mathbf{v_2} = 0 \).
2Step 2: Calculate the Dot Product
The dot product formula for \( \mathbf{v_1} \) and \( \mathbf{v_2} \) is: \[ (2c \cdot 3) + (-8 \cdot c) = 6c - 8c. \]
3Step 3: Set the Dot Product to Zero
Since \( \mathbf{v_1} \) and \( \mathbf{v_2} \) are orthogonal, set the result of the dot product to zero: \[ 6c - 8c = 0. \]
4Step 4: Solve for \( c \)
Simplifying the equation gives: \[ -2c = 0. \]Divide both sides by \(-2\) to solve for \( c \): \[ c = 0. \]
5Step 5: Confirm the Solution
Plug \( c = 0 \) back into the vector components to ensure orthogonality. If \( c = 0 \), \( \mathbf{v_1} = 0 \mathbf{i} - 8 \mathbf{j} \) and \( \mathbf{v_2} = 3 \mathbf{i} + 0 \mathbf{j} \), and the dot product is indeed 0.
Key Concepts
Dot Product and OrthogonalityUnderstanding Vector ComponentsSolving Equations for Orthogonality
Dot Product and Orthogonality
In the realm of vectors, the dot product is a key operation that allows us to measure various relationships between vectors.
Specifically, it helps determine if two vectors are orthogonal, meaning they form a right angle with each other. The dot product of two vectors, say \(\mathbf{v_1} = a\mathbf{i} + b\mathbf{j}\) and \(\mathbf{v_2} = c\mathbf{i} + d\mathbf{j}\), is calculated using the formula:
Specifically, it helps determine if two vectors are orthogonal, meaning they form a right angle with each other. The dot product of two vectors, say \(\mathbf{v_1} = a\mathbf{i} + b\mathbf{j}\) and \(\mathbf{v_2} = c\mathbf{i} + d\mathbf{j}\), is calculated using the formula:
- \(\mathbf{v_1} \cdot \mathbf{v_2} = ac + bd\)
Understanding Vector Components
Vectors are often defined in terms of their components along the axes of a coordinate system. For instance, in a 2D plane, a vector can be expressed in terms of its \( \mathbf{i} \) (horizontal) and \( \mathbf{j} \) (vertical) components. Each component represents how much the vector stretches along the respective axis.Consider the vectors \(2c \mathbf{i} - 8 \mathbf{j}\) and \(3 \mathbf{i} + c \mathbf{j}\) from the exercise:
- The vector \(2c \mathbf{i} - 8 \mathbf{j}\) has a horizontal component \(2c\) and a vertical component \(-8\).
- The vector \(3 \mathbf{i} + c \mathbf{j}\) has a horizontal component \(3\) and a vertical component \(c\).
Solving Equations for Orthogonality
To find when two vectors are orthogonal, you set their dot product equal to zero and solve for the unknown variable. In our exercise, the given vectors were \(2c \mathbf{i} - 8 \mathbf{j}\) and \(3 \mathbf{i} + c \mathbf{j}\). The equation derived from their dot product was:
Plugging \(c = 0\) back into the vector equations confirmed the vectors are orthogonal.
This step-by-step method illustrates how solving equations can help determine specific conditions for vector relationships such as orthogonality.
- \(6c - 8c = 0\)
- \(-2c = 0\)
Plugging \(c = 0\) back into the vector equations confirmed the vectors are orthogonal.
This step-by-step method illustrates how solving equations can help determine specific conditions for vector relationships such as orthogonality.
Other exercises in this chapter
Problem 20
Find the equation of the plane containing the line \(x=3 t, y=1+t, z=2 t\) and parallel to the intersection of the planes \(2 x-y+z=0\) and \(y+z+1=0\)
View solution Problem 20
Find the equation of the plane through the origin that is perpendicular to the \(x y\) -plane and the plane \(3 x-2 y+z=4\).
View solution Problem 20
An object weighing \(258.5\) pounds is held in equilibrium by two ropes that make angles of \(27.34^{\circ}\) and \(39.22^{\circ}\), respectively, with the vert
View solution Problem 21
The graph of an equation in \(x, y\), and \(z\) is symmetric with respect to the \(x y\) -plane if replacing \(z\) by \(-z\) results in an equivalent equation.
View solution