Math Problem Statement

Solve number 6 a

Дараах цэгүүд дээр оройтой ABC гурвалжны талбай болон аль нэг өнцгийн уртыг ол. (a) A(1, -1, 0), B(3, -3, 2), C(0, 2, 1)

Solution

To solve number 6 (a), we need to determine the area of the triangle ABCABC and find the length of one side of this triangle.

The points provided in option (a) are: A(1,−1,0),B(3,−3,2),C(0,2,1).A(1, -1, 0), \quad B(3, -3, 2), \quad C(0, 2, 1).

Step 1: Calculate Vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC}

First, find the vectors from AA to BB and from AA to CC.

  1. Vector AB→\overrightarrow{AB}: AB→=B−A=(3−1,−3−(−1),2−0)=(2,−2,2)\overrightarrow{AB} = B - A = (3 - 1, -3 - (-1), 2 - 0) = (2, -2, 2)

  2. Vector AC→\overrightarrow{AC}: AC→=C−A=(0−1,2−(−1),1−0)=(−1,3,1)\overrightarrow{AC} = C - A = (0 - 1, 2 - (-1), 1 - 0) = (-1, 3, 1)

Step 2: Calculate the Cross Product AB→×AC→\overrightarrow{AB} \times \overrightarrow{AC}

To find the area of the triangle, we need the magnitude of the cross product of AB→\overrightarrow{AB} and AC→\overrightarrow{AC}.

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 2 \\ -1 & 3 & 1 \\ \end{vmatrix}$$ Expanding this determinant: $$= \mathbf{i}((-2)(1) - (2)(3)) - \mathbf{j}((2)(1) - (2)(-1)) + \mathbf{k}((2)(3) - (-2)(-1))$$ $$= \mathbf{i}(-2 - 6) - \mathbf{j}(2 + 2) + \mathbf{k}(6 - 2)$$ $$= \mathbf{i}(-8) - \mathbf{j}(4) + \mathbf{k}(4)$$ $$= (-8, -4, 4)$$ ### Step 3: Find the Magnitude of the Cross Product The magnitude of $$\overrightarrow{AB} \times \overrightarrow{AC}$$ is: $$\sqrt{(-8)^2 + (-4)^2 + (4)^2} = \sqrt{64 + 16 + 16} = \sqrt{96} = 4\sqrt{6}$$ ### Step 4: Calculate the Area of Triangle $$ABC$$ The area $$A$$ of the triangle is half the magnitude of the cross product: $$\text{Area} = \frac{1}{2} \times 4\sqrt{6} = 2\sqrt{6}$$ ### Step 5: Calculate the Length of One Side (Example: $$AB$$) To find the length of $$AB$$, use the distance formula: $$AB = \sqrt{(3 - 1)^2 + (-3 + 1)^2 + (2 - 0)^2} = \sqrt{2^2 + (-2)^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$$ ### Answer for Part (a): - **Area of Triangle $$ABC$$:** $$2\sqrt{6}$$ - **Length of $$AB$$:** $$2\sqrt{3}$$ Would you like further explanation on any of these steps or help with another part? --- ### Additional Questions to Explore: 1. How would you find the area if the points were in a 2D plane instead of 3D? 2. What is the geometrical meaning of the cross product result in this context? 3. How can we find the perimeter of triangle $$ABC$$ using distances between all points? 4. How would the area calculation change if the coordinates of $$B$$ and $$C$$ were altered? 5. Can you verify the area using a different method, such as Heron's formula? #### Tip: The cross product of two vectors in 3D is useful for finding areas of parallelograms and triangles by relating it to the magnitude.

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Math Problem Analysis

Mathematical Concepts

Analytic Geometry
Vector Calculations
Cross Product
Distance Formula

Formulas

Vector AB = B - A
Vector AC = C - A
Cross Product AB x AC
Area of Triangle = 1/2 |AB x AC|
Distance formula for AB

Theorems

Cross Product for Area Calculation in 3D Geometry

Suitable Grade Level

Grades 10-12

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