Math Problem Statement

Please help me with this problem

Find a nonzero vector orthogonal to the plane through points P = (0, 0, -3), Q = (4, 2, 0), R = (3, 3, 1). Then find the area of triangle PQR.

Solution

To solve this problem, we’ll break it into two parts:

  1. Find a nonzero vector orthogonal to the plane containing the points P=(0,0,3)P = (0, 0, -3), Q=(4,2,0)Q = (4, 2, 0), and R=(3,3,1)R = (3, 3, 1).
  2. Calculate the area of triangle PQRPQR.

Step 1: Find a Nonzero Vector Orthogonal to the Plane

To find a vector orthogonal to the plane, we can use the cross product of two vectors that lie in the plane. We’ll construct two vectors in the plane from the given points:

  1. Vector PQ\overrightarrow{PQ}: PQ=QP=(40,20,0(3))=(4,2,3)\overrightarrow{PQ} = Q - P = (4 - 0, 2 - 0, 0 - (-3)) = (4, 2, 3)

  2. Vector PR\overrightarrow{PR}: PR=RP=(30,30,1(3))=(3,3,4)\overrightarrow{PR} = R - P = (3 - 0, 3 - 0, 1 - (-3)) = (3, 3, 4)

Now, calculate the cross product PQ×PR\overrightarrow{PQ} \times \overrightarrow{PR} to find a vector orthogonal to the plane:

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 2 & 3 \\ 3 & 3 & 4 \\ \end{vmatrix}$$ Expanding this determinant: $$= \hat{i}(2 \cdot 4 - 3 \cdot 3) - \hat{j}(4 \cdot 4 - 3 \cdot 3) + \hat{k}(4 \cdot 3 - 2 \cdot 3)$$ $$= \hat{i}(8 - 9) - \hat{j}(16 - 9) + \hat{k}(12 - 6)$$ $$= -\hat{i} - 7\hat{j} + 6\hat{k}$$ So, a nonzero vector orthogonal to the plane is: $$\vec{n} = (-1, -7, 6)$$ ### Step 2: Find the Area of Triangle $$PQR$$ The area of triangle $$PQR$$ is given by: $$\text{Area} = \frac{1}{2} \|\overrightarrow{PQ} \times \overrightarrow{PR}\|$$ First, find the magnitude of $$\overrightarrow{PQ} \times \overrightarrow{PR}$$: $$\|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \sqrt{(-1)^2 + (-7)^2 + 6^2}$$ $$= \sqrt{1 + 49 + 36} = \sqrt{86}$$ Thus, the area of triangle $$PQR$$ is: $$\text{Area} = \frac{1}{2} \sqrt{86}$$ ### Final Answers 1. **A nonzero vector orthogonal to the plane**: $$\vec{n} = (-1, -7, 6)$$. 2. **The area of triangle $$PQR$$**: $$\frac{1}{2} \sqrt{86}$$ (approximately 4.64 square units). Would you like further details on any specific step? --- Here are 5 additional questions for practice: 1. How would you find the equation of the plane containing points $$P$$, $$Q$$, and $$R$$? 2. What would happen if two of the points $$P$$, $$Q$$, and $$R$$ were collinear? 3. How do you interpret the cross product in terms of geometry? 4. Can a different orthogonal vector be found using scalar multiples of $$\vec{n}$$? 5. How would you verify that $$\vec{n}$$ is indeed orthogonal to the plane? **Tip:** Cross products are useful for finding perpendicular vectors in 3D geometry; remember that the magnitude of the cross product also gives the area of the parallelogram formed by the two vectors.

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Math Problem Analysis

Mathematical Concepts

Vectors
Cross Product
Plane Geometry
Vector Magnitude

Formulas

Cross product of two vectors: \( \overrightarrow{A} \times \overrightarrow{B} \)
Area of a triangle using vector cross product: \( \frac{1}{2} \|\overrightarrow{A} \times \overrightarrow{B}\| \)

Theorems

Properties of Cross Product
Area Formula for Triangles in 3D

Suitable Grade Level

Undergraduate Math (Calculus or Linear Algebra level)