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 \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \).
Step by step solution
01
Find vectors PQ and PR
Calculate vector \( \overrightarrow{PQ} \) from point \( P \) to point \( Q \) by subtracting the coordinates: \[ \overrightarrow{PQ} = Q - P = (3 - 2, -1 + 2, 2 - 1) = (1, 1, 1) \]Similarly, calculate vector \( \overrightarrow{PR} \) from point \( P \) to point \( R \):\[ \overrightarrow{PR} = R - P = (3 - 2, -1 + 2, 1 - 1) = (1, 1, 0) \]
02
Compute cross product of PQ and PR
Find the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) to get a vector perpendicular to the plane \(PQR\):\[\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 1 & 1 \1 & 1 & 0 \end{vmatrix}= \mathbf{i}(1 \cdot 0 - 1 \cdot 1) - \mathbf{j}(1 \cdot 0 - 1 \cdot 1) + \mathbf{k}(1 \cdot 1 - 1 \cdot 1)= -\mathbf{i} + \mathbf{j} + 0\mathbf{k}= (-1, 1, 0)\]
03
Find the magnitude of the cross product
Calculate the magnitude of \( \overrightarrow{PQ} \times \overrightarrow{PR} \):\[ \left| (-1, 1, 0) \right| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2} \]
04
Calculate the area of the triangle
The area \( A \) of triangle \(PQR\) is given by:\[ A = \frac{1}{2} \times \left| \overrightarrow{PQ} \times \overrightarrow{PR} \right| = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \]
05
Calculate the unit vector perpendicular to the plane
To find the unit vector, divide the cross product by its magnitude:\[ \text{Unit Vector} = \frac{(-1, 1, 0)}{\sqrt{2}} = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross Product
The cross product is a fundamental operation in vector calculus. It is primarily used to find a vector that is perpendicular to two given vectors in three-dimensional space. Unlike the dot product, which results in a scalar, the cross product results in a vector.
To calculate the cross product of two vectors, we use the determinant of a matrix. For example, given vectors \( \overrightarrow{PQ} = (1, 1, 1) \) and \( \overrightarrow{PR} = (1, 1, 0) \) in the exercise, the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) is computed as follows:
This vector is perpendicular to both \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \), and hence perpendicular to the plane formed by these vectors.
To calculate the cross product of two vectors, we use the determinant of a matrix. For example, given vectors \( \overrightarrow{PQ} = (1, 1, 1) \) and \( \overrightarrow{PR} = (1, 1, 0) \) in the exercise, the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) is computed as follows:
- Write the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row.
- Write the components of \( \overrightarrow{PQ} \) in the second row: \( 1, 1, 1 \).
- Write the components of \( \overrightarrow{PR} \) in the third row: \( 1, 1, 0 \).
This vector is perpendicular to both \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \), and hence perpendicular to the plane formed by these vectors.
Vectors
Vectors are essential in mathematics and physics, representing quantities having both magnitude and direction. In three-dimensional space, they are often represented as ordered triplets, such as \((x, y, z)\).
In the exercise, we calculate vectors using the points \(P(2, -2, 1)\), \(Q(3, -1, 2)\), and \(R(3, -1, 1)\). To find vector \( \overrightarrow{PQ} \), subtract the coordinates of \(P\) from \(Q\):
Vectors allow us to model and solve problems involving motion, forces, and other physical phenomena.
In the exercise, we calculate vectors using the points \(P(2, -2, 1)\), \(Q(3, -1, 2)\), and \(R(3, -1, 1)\). To find vector \( \overrightarrow{PQ} \), subtract the coordinates of \(P\) from \(Q\):
- \(Q_x - P_x = 3 - 2 = 1\)
- \(Q_y - P_y = -1 - (-2) = 1\)
- \(Q_z - P_z = 2 - 1 = 1\)
Vectors allow us to model and solve problems involving motion, forces, and other physical phenomena.
Unit Vector
A unit vector is a vector with a magnitude of one. It is used to indicate direction without concern for magnitude. To convert any vector into a unit vector, divide the vector by its magnitude.
In the exercise, we have the vector \((-1, 1, 0)\) from the cross product computation. To find the unit vector:
Unit vectors are extremely useful in physics for indicating directions, and in computer graphics for defining orientations.
In the exercise, we have the vector \((-1, 1, 0)\) from the cross product computation. To find the unit vector:
- First, compute the magnitude: \( \left| (-1, 1, 0) \right| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2} \).
- Then, divide each component of the vector by the magnitude: \((-1, 1, 0) / \sqrt{2} = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)\).
Unit vectors are extremely useful in physics for indicating directions, and in computer graphics for defining orientations.