 Suggested languages for you:

Europe

Problem 585

# Describe geometrically the solutions to $\mathrm{x}_{1}-2 \mathrm{x}_{2}+4 \mathrm{x}_{3}=1$ $$3 \mathrm{x}_{1}+\mathrm{x}_{2}-\mathrm{x}_{3}=0$$

Expert verified
The given system of linear equations represents two planes in 3-dimensional space, which are not parallel. They intersect in a line with the direction vector $$\langle-2, 13, 7 \rangle$$. The parametric equations of the intersection line are: $$x = 1 - 4t + \frac{13t - 3}{3}$$ $$y = \frac{13t - 3}{6}$$ $$z = t$$ These equations describe the geometric solution to the given system of linear equations.
See the step by step solution

## Step 1: Write the plane equations in the standard form

We'll write the given equation in the standard form (Ax + By + Cz = D) to make it easier to understand the geometric representations. The given equations are: 1. $$x_1 - 2x_2 + 4x_3 = 1$$ 2. $$3x_1 + x_2 - x_3 = 0$$ Now, let's rewrite these equations using x, y, and z instead of x1, x2, and x3: 1. $$x - 2y + 4z = 1$$ 2. $$3x + y - z = 0$$

## Step 2: Identify the normal vectors of the planes

The normal vector to a plane can be found from the coefficients of the variables. For Ax + By + Cz = D, the normal vector  is orthogonal to the plane. Normal vector for the first plane: $$<1, -2, 4>$$ Normal vector for the second plane: $$<3, 1, -1>$$

## Step 3: Check if the planes are parallel

Parallel planes never intersect. To check if the given planes are parallel, we need to check if their normal vectors are proportional. However, as we can see, the normal vectors are not proportional: $$\frac{1}{3} \neq \frac{-2}{1} \neq \frac{4}{-1}$$ Since the normal vectors are not proportional, the two planes are not parallel. Hence, they intersect in a line.

## Step 4: Find the direction vector of the intersection line

Since we know the two planes intersect, their intersection line's direction vector can be found as the cross product of the normal vectors of the planes. Let's find the cross product of the two normal vectors: $<1, -2, 4> \times <3, 1, -1> = <(-2)(-1) - (4)(1), (4)(3) - (1)(-1), (1)(1) - (-2)(3)>$ $<2 - 4, 12 + 1, 1 + 6> = <-2, 13, 7>$ So, the direction vector of the intersection line is: $$<-2, 13, 7>$$

## Step 5: Parametric equations of the line

To get the intersection line's parametric equations, rewrite the planes' equations with one variable as the parameter. Setting z as the parameter from the first equation: $$t = z$$ From equation 1: $$x = 1 - 4t + 2y$$ From equation 2: $$3x + y - t = 0$$ Now, let's substitute the first equation into the second equation: $$3(1 - 4t + 2y) + y - t = 0$$ Solve for y: $$-12t + 6y + 3 - t = 0$$ $$-13t + 6y + 3 = 0$$ $$y = \frac{13t - 3}{6}$$ Finally, we can rewrite x from equation 1 using y: $$x = 1 - 4t + 2(\frac{13t - 3}{6})$$ $$x = 1 - 4t + \frac{13t - 3}{3}$$ Thus, the parametric equations of the intersection line are: $$x = 1 - 4t + \frac{13t - 3}{3}$$ $$y = \frac{13t - 3}{6}$$ $$z = t$$ These equations provide the geometric description of the solutions to the given system of linear equations.

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features ## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner 