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).
Solution
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:
Area=21∣PQ×PR∣
where PQ and PR are vectors formed by the points.
Find vector PQ:PQ=Q−P=(1−2,0−(−2),−2−1)=(−1,2,−3)
Find vector PR:PR=R−P=(0−2,1−(−2),2−1)=(−2,3,1)
Compute the cross product PQ×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?
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**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.
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Math Problem Analysis
Mathematical Concepts
Vectors Cross Product Normalization 3D Geometry Unit Vector
Formulas
Area of triangle in 3D space = (1/2) * |PQ × PR| Cross Product of vectors PQ and PR Magnitude of vector: |v| = sqrt(x^2 + y^2 + z^2) Normalization: unit vector = (1 / |v|) * v
Theorems
Cross Product Theorem: The cross product of two vectors in 3D gives a vector perpendicular to the plane formed by the vectors. Pythagoras Theorem for 3D Magnitude Calculation