Math Problem Statement
Please help me with this problem
Solution
To solve this problem, we’ll break it into two parts:
- Find a nonzero vector orthogonal to the plane containing the points , , and .
- Calculate the area of triangle .
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:
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Vector :
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Vector :
Now, calculate the cross product 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.Ask a new question for Free
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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)
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