https://file.aiquickdraw.com/mathbot/file/1729500943472m95rfj78.jpg

Let's break this problem down.

### Part a: Find the area of the triangle determined by points \( P(2, -2, 1) \), \( Q(1, 0, -2) \), and \( R(0, 1, 2) \).

The area of a triangle in 3D space is given by:
\[
\text{Area} = \frac{1}{2} \left| \mathbf{PQ} \times \mathbf{PR} \right|
\]
where \( \mathbf{PQ} \) and \( \mathbf{PR} \) are vectors formed by the points.

1. **Find vector \( \mathbf{PQ} \):**
\[
\mathbf{PQ} = Q - P = (1 - 2, 0 - (-2), -2 - 1) = (-1, 2, -3)
\]

2. **Find vector \( \mathbf{PR} \):**
\[
\mathbf{PR} = R - P = (0 - 2, 1 - (-2), 2 - 1) = (-2, 3, 1)
\]

3. **Compute the cross product \( \mathbf{PQ} \times \mathbf{PR} \):**

\[
\mathbf{PQ} \times \mathbf{PR} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 2 & -3 \\
-2 & 3 & 1
\end{vmatrix}
\]

\[
= \mathbf{i} \left( 2 \cdot 1 - (-3) \cdot 3 \right) - \mathbf{j} \left( (-1) \cdot 1 - (-3) \cdot (-2) \right) + \mathbf{k} \left( (-1) \cdot 3 - 2 \cdot (-2) \right)
\]
\[
= \mathbf{i}(2 + 9) - \mathbf{j}(-1 - 6) + \mathbf{k}(-3 + 4)
\]
\[
= \mathbf{i}(11) - \mathbf{j}(-7) + \mathbf{k}(1)
\]
\[
= 11\mathbf{i} + 7\mathbf{j} + \mathbf{k}
\]

4. **Find the magnitude of the cross product:**
\[
\left| \mathbf{PQ} \times \mathbf{PR} \right| = \sqrt{11^2 + 7^2 + 1^2} = \sqrt{121 + 49 + 1} = \sqrt{171}
\]

5. **Area of the triangle:**
\[
\text{Area} = \frac{1}{2} \times \sqrt{171} = \frac{\sqrt{171}}{2}
\]

### Part b: Find a unit vector perpendicular to the plane PQR

A unit vector perpendicular to the plane is in the same direction as the cross product \( \mathbf{PQ} \times \mathbf{PR} \), but normalized.

The normal vector we calculated is \( 11\mathbf{i} + 7\mathbf{j} + \mathbf{k} \).

1. **Magnitude of the normal vector:**
\[
\left| \mathbf{n} \right| = \sqrt{11^2 + 7^2 + 1^2} = \sqrt{171}
\]

2. **Normalize the vector:**
The unit vector is:
\[
\mathbf{u} = \frac{1}{\sqrt{171}} (11\mathbf{i} + 7\mathbf{j} + \mathbf{k})
\]

Now, comparing this to the options provided:

- **Option A** gives:**
\[
\frac{1}{3\sqrt{19}} (11\mathbf{i} + 7\mathbf{j} + \mathbf{k})
\]
This matches the form we calculated, but it needs to be multiplied by \( \frac{1}{\sqrt{171}} \), and \( \sqrt{171} \) can be simplified to \( 3\sqrt{19} \), which makes **Option A** correct.

### Final Answers:
- **Part a (Area)**: \( \frac{\sqrt{171}}{2} \)
- **Part b (Unit Vector)**: **Option A**

Do you want further details or have any other questions?

---

**Related questions:**
1. How do you calculate the cross product in 3D geometry?
2. What is the significance of the magnitude of the cross product in determining the area?
3. How do you find the normal vector to a plane in 3D space?
4. What is the geometric interpretation of a unit vector?
5. How does normalization affect a vector?

**Tip**: Remember that the cross product of two vectors gives a vector perpendicular to both, making it useful for finding normal vectors in geometry.

a) Find the area of the triangle determined by points P(2, -2, 1), Q(1, 0, -2), and R(0, 1, 2).
b) Find a unit vector perpendicular to plane PQR. The options provided are:
A) (1 / (3√19))(11i + 7j + k)
B) (1 / √2)(i + j)
C) (1 / (3√19))(7i + 11j + k)
D) (1 / √6)(2i + j + k).

This problem explores how to calculate the area of a triangle in 3D space formed by points P(2, -2, 1), Q(1, 0, -2), and R(0, 1, 2). It also examines how to find a unit vector perpendicular to the plane using the cross product of vectors PQ and PR.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation