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Chapter 11: Problem 19

For what numbers \(c\) are \(\langle c, 6\rangle\) and \(\langle c,-4\rangle\) orthogonal?

Short Answer

Expert verified
The numbers are \( c = 2\sqrt{6} \) and \( c = -2\sqrt{6} \).

Step by step solution

01

Understanding Orthogonal Vectors

Two vectors are orthogonal if their dot product is zero. Thus, we need to calculate the dot product of the vectors \( \langle c, 6 \rangle \) and \( \langle c, -4 \rangle \).
02

Setup the Dot Product

The dot product formula for vectors \( \langle a, b \rangle \) and \( \langle x, y \rangle \) is \( a \cdot x + b \cdot y \). Apply this formula to our vectors: \( c \cdot c + 6 \cdot (-4) = c^2 - 24 \).
03

Set the Dot Product to Zero

To find when the vectors are orthogonal, set the dot product equal to zero: \( c^2 - 24 = 0 \).
04

Solve the Equation for c

Solve the equation \( c^2 - 24 = 0 \) for \( c \). Add 24 to both sides: \( c^2 = 24 \). Then, taking the square root of both sides gives \( c = \sqrt{24} \) or \( c = -\sqrt{24} \). Simplify \( \sqrt{24} \) to \( 2\sqrt{6} \), so \( c = 2\sqrt{6} \) or \( c = -2\sqrt{6} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product is a fundamental operation in vector calculus. It simplifies calculations involving vectors. Here's how you can understand it:
  • To compute the dot product of two vectors, you multiply their corresponding components and add the results. For vectors \( \langle a, b \rangle \) and \( \langle x, y \rangle \), the dot product is \( a \cdot x + b \cdot y \).
  • If two vectors are orthogonal, their dot product is zero. This means they are perpendicular to each other.
In our exercise, we have vectors \( \langle c, 6 \rangle \) and \( \langle c, -4 \rangle \). Their dot product becomes \( c \cdot c + 6 \cdot (-4) \). This simplifies to \( c^2 - 24 \). For the vectors to be orthogonal, set this expression to zero.
Vector Calculus
Vector calculus deals with the differentiation and integration of vector fields. It's a powerful mathematical tool used to describe physical phenomena in various fields like physics and engineering.
  • Key operations include calculating the dot product, as seen in our exercise, and the cross product, which finds a vector perpendicular to two given vectors.
  • Vector calculus is helpful for modeling forces, fields, and motion in a multidimensional space.
In the problem, we use vector operations to find when vectors are orthogonal. This aspect of vector calculus allows us to determine angles and relationships between vectors, enhancing our understanding of geometry in space.
Quadratic Equation
A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). It's one of the basic equations in algebra and is solved using various methods:
  • Factoring, when possible, is the simplest approach.
  • The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) always provides a solution.
  • Completing the square is another way to solve quadratics, giving insights into their properties.
In the exercise, the dot product led to the quadratic equation \( c^2 - 24 = 0 \). Solving it involves isolating \( c^2 \) and taking the square root of both sides, yielding \( c = \pm 2\sqrt{6} \). This solution reveals the values for which the vectors are orthogonal.

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