a. Find the area of the triangle determined by the points​ P, Q, and R.
b. Find a unit vector perpendicular to plane PQR.
Upper P left parenthesis 2 comma negative 2 comma 1 right parenthesis​,     Upper Q left parenthesis 1 comma 0 comma negative 2 right parenthesis​,     Upper R left parenthesis 0 comma 1 comma 2 right parenthesis

To solve the problems related to the points \( P(2, -2, 1) \), \( Q(1, 0, -2) \), and \( R(0, 1, 2) \), we will go through the following steps:

### Part (a): Finding the Area of the Triangle Determined by Points \( P \), \( Q \), and \( R \)

The area of the triangle can be found using the formula:
\[
\text{Area} = \frac{1}{2} \left\| \overrightarrow{PQ} \times \overrightarrow{PR} \right\|
\]
where:
\[
\overrightarrow{PQ} = Q - P = (1-2, 0-(-2), -2-1) = (-1, 2, -3)
\]
\[
\overrightarrow{PR} = R - P = (0-2, 1-(-2), 2-1) = (-2, 3, 1)
\]

Now, compute the cross product \(\overrightarrow{PQ} \times \overrightarrow{PR}\):
\[
\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 2 & -3 \\
-2 & 3 & 1 \\
\end{vmatrix}
\]
Expanding this determinant:
\[
\overrightarrow{PQ} \times \overrightarrow{PR} = \mathbf{i}(2 \times 1 - (-3) \times 3) - \mathbf{j}((-1) \times 1 - (-3) \times (-2)) + \mathbf{k}((-1) \times 3 - 2 \times (-2))
\]
\[
= \mathbf{i}(2 + 9) - \mathbf{j}(-1 - 6) + \mathbf{k}(-3 + 4)
\]
\[
= \mathbf{i}(11) - \mathbf{j}(-7) + \mathbf{k}(1)
\]
\[
\overrightarrow{PQ} \times \overrightarrow{PR} = (11, 7, 1)
\]

The magnitude of the cross product is:
\[
\left\| \overrightarrow{PQ} \times \overrightarrow{PR} \right\| = \sqrt{11^2 + 7^2 + 1^2} = \sqrt{121 + 49 + 1} = \sqrt{171}
\]

Therefore, the area of the triangle is:
\[
\text{Area} = \frac{1}{2} \times \sqrt{171} \approx \frac{1}{2} \times 13.08 = 6.54
\]

### Part (b): Finding a Unit Vector Perpendicular to Plane \( PQR \)

A vector perpendicular to the plane can be found by normalizing the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \).

The unit vector is:
\[
\text{Unit vector} = \frac{\overrightarrow{PQ} \times \overrightarrow{PR}}{\left\| \overrightarrow{PQ} \times \overrightarrow{PR} \right\|} = \frac{(11, 7, 1)}{\sqrt{171}}
\]
\[
= \left( \frac{11}{\sqrt{171}}, \frac{7}{\sqrt{171}}, \frac{1}{\sqrt{171}} \right)
\]

So, the unit vector perpendicular to the plane \( PQR \) is:
\[
\left( \frac{11}{\sqrt{171}}, \frac{7}{\sqrt{171}}, \frac{1}{\sqrt{171}} \right)
\]

### Summary:
- **Area of the triangle:** \( 6.54 \) square units.
- **Unit vector perpendicular to the plane:** \( \left( \frac{11}{\sqrt{171}}, \frac{7}{\sqrt{171}}, \frac{1}{\sqrt{171}} \right) \).

Would you like more details on any part of the solution? Here are some related questions:

1. How is the cross product calculated for vectors in three dimensions?
2. What are the properties of the cross product in vector analysis?
3. How can the magnitude of a vector be determined?
4. How does the area of a triangle relate to the cross product?
5. What are the applications of unit vectors in physics and engineering?

**Tip:** Remember, the cross product of two vectors gives a vector that is perpendicular to both, which is crucial in determining the normal to a plane.

Learn how to calculate the area of a triangle defined by points P(2, -2, 1), Q(1, 0, -2), and R(0, 1, 2) using vectors and the cross product. Also, find a unit vector perpendicular to the plane formed by these points. Suitable for advanced high school or college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation