Problem 17 a. Find the area of the triangle... [FREE SOLUTION]

Chapter 11: Problem 17

a. Find the area of the triangle determined by the points $P, Q$ and $R$ b. Find a unit vector perpendicular to plane $P Q R$ $$P(2,-2,1), \quad Q(3,-1,2), \quad R(3,-1,1)$$

Short Answer

Expert verified

a. The area is $ \frac{\sqrt{2}}{2} $. b. The unit vector is $ \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle $.

Step by step solution

Vector Calculation

To calculate the area of the triangle and a vector perpendicular to the plane, we first need the vectors along two sides of the triangle. Calculate $ \vec{PQ} $ and $ \vec{PR} $. \[\vec{PQ} = Q - P = \langle 3 - 2, -1 + 2, 2 - 1 \rangle = \langle 1, 1, 1 \rangle\]\[\vec{PR} = R - P = \langle 3 - 2, -1 + 2, 1 - 1 \rangle = \langle 1, 1, 0 \rangle\]

Finding the Cross Product

The cross product $ \vec{PQ} \times \vec{PR} $ gives a vector perpendicular to the plane of the triangle. Calculate it using the determinant.\[\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} = \mathbf{i}(1\cdot0 - 1\cdot1) - \mathbf{j}(1\cdot0 - 1\cdot1) + \mathbf{k}(1\cdot1 - 1\cdot1) = \langle -1, 1, 0 \rangle\]

Calculate the Area of the Triangle

The area of triangle $ PQR $ is half the magnitude of the cross product $ \vec{PQ} \times \vec{PR} $. Calculate the magnitude.\[|\vec{PQ} \times \vec{PR}| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2}\]Thus, the area is\[\text{Area} = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2}\]

Finding the Unit Vector

To find the unit vector, divide the cross product by its magnitude. The magnitude is $ \sqrt{2} $ as calculated above.The unit vector $ \mathbf{n} $ is:\[\mathbf{n} = \frac{1}{\sqrt{2}} \langle -1, 1, 0 \rangle = \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle\]

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cross Product

The cross product is a vector operation that allows us to find a perpendicular vector to a plane defined by two vectors. This is immensely useful in 3D space, such as when dealing with physics or engineering problems. Given two vectors, $ \vec{a} $ and $ \vec{b} $, the cross product $ \vec{a} \times \vec{b} $ is another vector:

This resulting vector is perpendicular to both $ \vec{a} $ and $ \vec{b} $.
The magnitude of the cross product represents the area of the parallelogram that the vectors span.
The cross product is given by a determinant calculation, often using the standard unit vectors $ \mathbf{i}, \mathbf{j}, \mathbf{k} $.

For example, using the vectors $ \vec{PQ} = \langle 1, 1, 1 \rangle $ and $ \vec{PR} = \langle 1, 1, 0 \rangle $, we calculated the cross product to be $ \langle -1, 1, 0 \rangle $. This perpendicular vector is crucial for subsequent calculations.

Triangle Area Calculation

The area of a triangle can be calculated using the cross product if the triangle is defined in three-dimensional space by two vectors originating from the same point. Here's how it works:

When vectors $ \vec{PQ} $ and $ \vec{PR} $ are used to span a space, the magnitude of their cross product correlates to the area of a parallelogram.
The area of the triangle is half the area of this parallelogram, because a triangle is essentially half of a parallelogram.

In the given problem, after finding the cross product, we determined its magnitude $ \sqrt{2} $. Therefore, the area of triangle $ PQR $ was calculated as $ \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} $.

Perpendicular Vector

A perpendicular vector is one that forms a 90-degree angle with another vector or a plane. Identify it using the cross product in vector calculus. In three-dimensional geometry:

The cross product of two non-parallel vectors results in a vector perpendicular to both.
This perpendicular vector is significant, for instance, in defining normal lines to planes or surfaces.

In our exercise, the vector $ \langle -1, 1, 0 \rangle $ is perpendicular to the plane formed by the triangle $ PQR $. Knowing perpendicular vectors helps in formulating the equation of planes and various computer graphics applications.

Unit Vector

A unit vector has a magnitude of 1 and is used to indicate direction without regard to scale. You compute it by dividing a vector by its own magnitude. This process normalizes the vector while retaining its direction. Here's the step-by-step method to find a unit vector:

Calculate the magnitude $|\vec{v}|$ of your vector $\vec{v}$.
Divide each component of the vector by its magnitude: $\mathbf{u} = \frac{\vec{v}}{|\vec{v}|}$.

For the perpendicular vector $ \langle -1, 1, 0 \rangle $, the magnitude was $\sqrt{2}$. So, to find the unit vector $\mathbf{n}$, we divided each component by $\sqrt{2}$, resulting in the unit vector $ \langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \rangle $. This technique is valuable in fields like physics and engineering, where directionality matters more than magnitude.

91影视

a. Find the area of the triangle determined by the points \(P, Q\) and \(R\) b. Find a unit vector perpendicular to plane \(P Q R\) $$P(2,-2,1), \quad Q(3,-1,2), \quad R(3,-1,1)$$

Short Answer

Step by step solution

Vector Calculation

Finding the Cross Product

Calculate the Area of the Triangle

Finding the Unit Vector

Key Concepts

Cross Product

Triangle Area Calculation

Perpendicular Vector

Unit Vector

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Theoretical and Mathematical Physics

Mechanics Maths

Probability and Statistics

Statistics

Decision Maths

Study anywhere. Anytime. Across all devices.