Problem 10

Question

$$ \text { Let } A=(1,1,3) \text { and } B=(2,-1,1) \text { . Can you find a point } p \text { such that } p \cdot A=0 \text { and } p \cdot B=0 \text { ? } $$

Step-by-Step Solution

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Answer

The point $p$ that satisfies the conditions that $p \cdot A=0$ and $p \cdot B=0$ is $p = (2, 7, -3)$.

1Step 1: Understand the problem and write down the equations.

The given expressions $p \cdot A=0$ and $p \cdot B=0$ translate to the following two equations: \[p \cdot (1,1,3) = 0\] \[p \cdot (2,-1,1) = 0\] assuming $p = (x, y, z)$, the above equations become: $x+y+3z=0$ and $2x-y+z=0$. These are the main equations to solve in order to find $p$.

2Step 2: Use substitution method.

Now, solve these two equations for $x, y, z$, the components of vector $p$. You can start by expressing $y$ from the equation $x+y+3z=0$ as $y= -x-3z$, and substituting this in the second equation $2x-y+z=0$.

3Step 3: Solve the two equations.

Substituting $y = -x-3z$ into the equation $2x-y+z=0$, the equation becomes: \[2x -(-x-3z) + z = 0\] Simplifying it, we get: \[3x + 2z = 0\]. From here, any real numbers could been chosen for $x$ and $z$. Choose, for example, $x = 2$ and $z = -3$. Deducing from these into the equation $y= -x-3z$, we get $y = 7$.

4Step 4: Construct the desired point $p$.

The components $x, y, z$ of the vector $p$ are now known, with $x = 2$, $y = 7$, and $z = -3$. Thus, the point $p$ that satisfies the given conditions is $p = (2, 7, -3)$.

Key Concepts

Dot ProductSystem of EquationsVector Components

Dot Product

The dot product is a fundamental concept in linear algebra that allows us to multiply two vectors to get a scalar. Imagine it as a way to measure how much one vector extends in the direction of another. The mathematical definition of the dot product for two vectors $ \textbf{u} = (u_1, u_2, u_3) $ and $ \textbf{v} = (v_1, v_2, v_3) $ is given by:

$ \textbf{u} \cdot \textbf{v} = u_1v_1 + u_2v_2 + u_3v_3 $

The exercise shows a practical application of the dot product, where point $ p $ is sought such that its dot product with vectors $ A $ and $ B $ both equal zero. This condition, $ p \cdot A = 0 $ and $ p \cdot B = 0 $, indicates that $ p $ is orthogonal (perpendicular) to both $ A $ and $ B $. When two vectors are orthogonal, their dot product is zero, which is a key property used here to derive the necessary equations.

System of Equations

Solving systems of equations is a method used to find values that satisfy multiple conditions simultaneously. In the given exercise, you are tasked with finding the vector $ p = (x, y, z) $ such that both dot products $ p \cdot A = 0 $ and $ p \cdot B = 0 $ hold true. These conditions translate into two linear equations as follows:

$ x + y + 3z = 0 $
$ 2x - y + z = 0 $

The challenge now is to solve this system of equations. A common approach is the substitution method, where you express one variable in terms of the others and substitute it into the other equation. This reduces the complexity and helps find a consistent solution, as demonstrated step-by-step in the exercise. The goal is to find a set of values for $ x, y, $ and $ z $ that makes both these equations true.

Vector Components

Vectors are mathematical objects with both magnitude and direction, expressed in terms of components—individual elements representing each dimension in a vector space. For a vector $ p = (x, y, z) $, each component corresponds to a coordinate along the respective axis.Understanding vector components is crucial for operations involving vectors, such as the dot product and solving systems of equations. Each component affects the calculations you perform with vectors. In this exercise, the components $ x, y, $ and $ z $ of vector $ p $ determine the outcome of the conditions $ p \cdot A = 0 $ and $ p \cdot B = 0 $.Working with vector components involves finding values that satisfy the given conditions, such as solving the linear equations formed by setting the dot products to zero. This knowledge adds up, enabling one to interpret the orientation and position of vectors relative to each other, as seen with $ A $ and $ B $ in the exercise.

Problem 9

Problem 10

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