Math Problem Statement

Solve

Find the area of the triangle determined by the points P, Q, and R. Find a unit vector perpendicular to plane PQR. Points: P(-2,2,1), Q(1,0,2), R(0,1,-2).

Solution

Problem Breakdown and Solution:

We are tasked with the following:

  1. Find the area of the triangle determined by the points P(−2,2,1)P(-2, 2, 1), Q(1,0,2)Q(1, 0, 2), and R(0,1,−2)R(0, 1, -2).
  2. Find a unit vector perpendicular to the plane PQRPQR.

Step 1: Vectors in the Plane

First, we find two vectors in the plane determined by the three points PP, QQ, and RR:

  • Vector PQ→=Q−P\overrightarrow{PQ} = Q - P: PQ→=(1−(−2),0−2,2−1)=(3,−2,1)\overrightarrow{PQ} = (1 - (-2), 0 - 2, 2 - 1) = (3, -2, 1)

  • Vector PR→=R−P\overrightarrow{PR} = R - P: PR→=(0−(−2),1−2,−2−1)=(2,−1,−3)\overrightarrow{PR} = (0 - (-2), 1 - 2, -2 - 1) = (2, -1, -3)

Step 2: Cross Product of PQ→\overrightarrow{PQ} and PR→\overrightarrow{PR}

The cross product PQ→×PR→\overrightarrow{PQ} \times \overrightarrow{PR} gives a vector perpendicular to the plane. Calculate it using the determinant formula:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 1 \\ 2 & -1 & -3 \end{vmatrix}$$ Expanding the determinant: $$\overrightarrow{PQ} \times \overrightarrow{PR} = \mathbf{i} \begin{vmatrix} -2 & 1 \\ -1 & -3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 1 \\ 2 & -3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & -2 \\ 2 & -1 \end{vmatrix}$$ 1. Compute each minor determinant: - $$\begin{vmatrix} -2 & 1 \\ -1 & -3 \end{vmatrix} = (-2)(-3) - (1)(-1) = 6 + 1 = 7$$ - $$\begin{vmatrix} 3 & 1 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (1)(2) = -9 - 2 = -11$$ - $$\begin{vmatrix} 3 & -2 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (-2)(2) = -3 + 4 = 1$$ 2. Substitute back: $$\overrightarrow{PQ} \times \overrightarrow{PR} = 7\mathbf{i} - (-11)\mathbf{j} + 1\mathbf{k}$$ $$\overrightarrow{PQ} \times \overrightarrow{PR} = 7\mathbf{i} + 11\mathbf{j} + \mathbf{k}$$ --- ### Step 3: Area of the Triangle The magnitude of the cross product is: $$\|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \sqrt{7^2 + 11^2 + 1^2} = \sqrt{49 + 121 + 1} = \sqrt{171}$$ The area of the triangle is half the magnitude of the cross product: $$\text{Area} = \frac{1}{2} \|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \frac{1}{2} \sqrt{171}$$ --- ### Step 4: Unit Vector Perpendicular to the Plane To find the unit vector, divide the cross product by its magnitude: $$\text{Unit Vector} = \frac{\overrightarrow{PQ} \times \overrightarrow{PR}}{\|\overrightarrow{PQ} \times \overrightarrow{PR}\|}$$ Substitute: $$\text{Unit Vector} = \frac{7\mathbf{i} + 11\mathbf{j} + \mathbf{k}}{\sqrt{171}}$$ This simplifies to: $$\text{Unit Vector} = \frac{7}{\sqrt{171}}\mathbf{i} + \frac{11}{\sqrt{171}}\mathbf{j} + \frac{1}{\sqrt{171}}\mathbf{k}$$ --- ### Final Answers: 1. **Area of the Triangle**: $$\frac{\sqrt{171}}{2}$$ 2. **Unit Vector Perpendicular to the Plane**: $$\frac{1}{\sqrt{171}}(7\mathbf{i} + 11\mathbf{j} + \mathbf{k})$$. Matching the options in the question, the correct choices are: - **Area**: $$\frac{\sqrt{171}}{2}$$ - **Unit Vector**: $$\frac{1}{\sqrt{171}}(7\mathbf{i} + 11\mathbf{j} + \mathbf{k})$$, corresponding to Option A. --- Do you have any questions or want further clarification? ### 5 Related Questions: 1. How is the cross product used to determine the perpendicularity of vectors? 2. Why is the magnitude of the cross product used to calculate the area of a triangle? 3. What is the geometric meaning of a unit vector? 4. How would the result change if the coordinates of one point were shifted? 5. Can this method be extended to compute areas of polygons in 3D? ### Tip: When working with cross products, always double-check determinant calculations for sign errors!

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Math Problem Analysis

Mathematical Concepts

Vectors
Cross Product
Unit Vector
Triangle Area in 3D

Formulas

Cross Product Formula: a × b = |i j k|
Triangle Area: Area = 1/2 * ||a × b||
Unit Vector: u = v / ||v||

Theorems

Properties of the Cross Product

Suitable Grade Level

College-level Mathematics

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