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Chapter 14: Problem 18

$$\text { Find } x \text { and } y \text { such that }\left(\begin{array}{l} 2 \\ x \\ y \end{array}\right) \text { is perpendicular to both }\left(\begin{array}{r} 3 \\ 1 \\ -1 \end{array}\right) \text { and }\left(\begin{array}{r} 4 \\ -1 \\ 2 \end{array}\right)$$

Short Answer

Expert verified
\( x = -20 \) and \( y = -14 \).

Step by step solution

01

Understand Perpendicularity

For a vector \( \mathbf{v} \) to be perpendicular to another vector \( \mathbf{u} \), their dot product must equal zero. This means \( \mathbf{u} \cdot \mathbf{v} = 0 \). We need to apply this to both given vectors.
02

Apply Dot Product Condition to First Vector

Find the dot product of \( \begin{pmatrix} 2 \ x \ y \end{pmatrix} \) and \( \begin{pmatrix} 3 \ 1 \ -1 \end{pmatrix} \):\[2 \cdot 3 + x \cdot 1 + y \cdot (-1) = 6 + x - y = 0\]This simplifies to the equation: \( x - y = -6 \).
03

Apply Dot Product Condition to Second Vector

Find the dot product of \( \begin{pmatrix} 2 \ x \ y \end{pmatrix} \) and \( \begin{pmatrix} 4 \ -1 \ 2 \end{pmatrix} \):\[2 \cdot 4 + x \cdot (-1) + y \cdot 2 = 8 - x + 2y = 0\]This simplifies to the equation: \( -x + 2y = -8 \).
04

Solve the System of Equations

Now solve the simultaneous equations:1. \( x - y = -6 \)2. \( -x + 2y = -8 \)Add the two equations to eliminate \( x \):\[(x - y) + (-x + 2y) = -6 - 8\]\[y = -14\]Plug \( y = -14 \) into the first equation:\[x - (-14) = -6 \x + 14 = -6 \x = -20\]
05

Verify the Solution

Substitute \( x = -20 \) and \( y = -14 \) back into both initial dot product equations to verify.For \( x - y = -6 \):\[-20 - (-14) = -6 \] (True)For \( -x + 2y = -8 \):\[-(-20) + 2(-14) = 20 - 28 = -8 \] (True)Both equations are satisfied, so \( x = -20 \) and \( y = -14 \) are correct.

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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 concept in vector mathematics that serves as a tool for understanding the relationship between two vectors. It is calculated by multiplying the corresponding components of the vectors and summing the results. If you have vectors \( \mathbf{a} = \begin{pmatrix} a_1 \ a_2 \ a_3 \end{pmatrix} \) and \( \mathbf{b} = \begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix} \), the dot product is expressed as:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
The dot product can tell you a lot about the vectors:
  • If the dot product is zero, the vectors are perpendicular, meaning they meet at a right angle.
  • If the dot product is positive, the vectors point in a generally similar direction.
  • If the dot product is negative, they point in opposite directions.
Understanding dot products is crucial for solving problems involving perpendicular vectors, as it defines when two vectors in space do not influence each other directionally.
Perpendicular Vectors
When two vectors are perpendicular, they intersect at a 90-degree angle. This relationship is mathematically captured by the condition that their dot product equals zero. Let's take two vectors, \( \mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ u_3 \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix} \). They are perpendicular if:
  • \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 = 0 \)
This concept is widely used:
  • In physics, for determining orthogonality in force directions.
  • In engineering, for ensuring stability in structures.
  • In computer graphics, for calculating normals to surfaces which affect lighting and shading.
In the given exercise, understanding the perpendicularity principle is key to setting up the dot product equations that lead us to find the values of \( x \) and \( y \). This forms the basis for the system of equations that needs to be solved.
System of Equations
A system of equations is a set of equations with multiple variables that you solve together. Each equation represents a constraint on the variables, and the solution satisfies all conditions simultaneously. Consider two linear equations:
  • \( ax + by = c \)
  • \( dx + ey = f \)
To solve these, we often use methods such as:
  • Substitution: Solve one equation for one variable and substitute into the other.
  • Elimination: Add or subtract equations to eliminate one variable, simplifying the system to a single equation with one unknown.
  • Graphical Method: Plotting each equation on a graph helps visualize where solutions might lie.
In the exercise, we utilized the elimination method. By adding equations derived from the dot product conditions, we isolated one variable, making it easier to find its value and subsequently solve for the other variable. The solution \( (x, y) = (-20, -14) \) satisfies both original equations, verifying the correct configuration of perpendicularity for the vectors.

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Most popular questions from this chapter

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