Math Problem Statement

Докажите, что четыре точки A(1, 2, -1), B(0, 1, 5), C(-1, 2, 1) и D(2, 1, 3) лежат в одной плоскости.

Solution

To prove that four points A(1,2,−1)A(1, 2, -1), B(0,1,5)B(0, 1, 5), C(−1,2,1)C(-1, 2, 1), and D(2,1,3)D(2, 1, 3) lie in the same plane, we need to determine if the vectors formed by these points are coplanar. To do this, we can calculate the scalar triple product of three vectors formed from the given points. If the scalar triple product is zero, then the points are coplanar.

Step 1: Find the vectors

We first form vectors AB\mathbf{AB}, AC\mathbf{AC}, and AD\mathbf{AD}:

  • AB=B−A=(0−1,1−2,5−(−1))=(−1,−1,6)\mathbf{AB} = B - A = (0-1, 1-2, 5-(-1)) = (-1, -1, 6)
  • AC=C−A=(−1−1,2−2,1−(−1))=(−2,0,2)\mathbf{AC} = C - A = (-1-1, 2-2, 1-(-1)) = (-2, 0, 2)
  • AD=D−A=(2−1,1−2,3−(−1))=(1,−1,4)\mathbf{AD} = D - A = (2-1, 1-2, 3-(-1)) = (1, -1, 4)

Thus: AB=(−1,−1,6),AC=(−2,0,2),AD=(1,−1,4)\mathbf{AB} = (-1, -1, 6), \quad \mathbf{AC} = (-2, 0, 2), \quad \mathbf{AD} = (1, -1, 4)

Step 2: Compute the scalar triple product

The scalar triple product of vectors u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} is given by: u⋅(v×w)\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) Here, we compute AB⋅(AC×AD)\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}).

2.1: Calculate the cross product AC×AD\mathbf{AC} \times \mathbf{AD}

Using the determinant formula for the cross product:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}$$ For $$\mathbf{AC} = (-2, 0, 2)$$ and $$\mathbf{AD} = (1, -1, 4)$$: $$\mathbf{AC} \times \mathbf{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 2 \\ 1 & -1 & 4 \end{vmatrix}$$ Expanding the determinant: $$\mathbf{AC} \times \mathbf{AD} = \mathbf{i} \begin{vmatrix} 0 & 2 \\ -1 & 4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -2 & 2 \\ 1 & 4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -2 & 0 \\ 1 & -1 \end{vmatrix}$$ Simplify each minor: - $$\begin{vmatrix} 0 & 2 \\ -1 & 4 \end{vmatrix} = (0)(4) - (-1)(2) = 2$$ - $$\begin{vmatrix} -2 & 2 \\ 1 & 4 \end{vmatrix} = (-2)(4) - (1)(2) = -8 - 2 = -10$$ - $$\begin{vmatrix} -2 & 0 \\ 1 & -1 \end{vmatrix} = (-2)(-1) - (0)(1) = 2$$ Thus: $$\mathbf{AC} \times \mathbf{AD} = (2)\mathbf{i} - (-10)\mathbf{j} + (2)\mathbf{k} = (2, 10, 2)$$ --- #### 2.2: Compute the dot product $$\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD})$$ Now, calculate the dot product of $$\mathbf{AB} = (-1, -1, 6)$$ and $$\mathbf{AC} \times \mathbf{AD} = (2, 10, 2)$$: $$\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = (-1)(2) + (-1)(10) + (6)(2)$$ Simplify: $$\mathbf{AB} \cdot (\mathbf{AC} \times \mathbf{AD}) = -2 - 10 + 12 = 0$$ --- ### Step 3: Conclusion Since the scalar triple product equals zero, the vectors $$\mathbf{AB}$$, $$\mathbf{AC}$$, and $$\mathbf{AD}$$ are **coplanar**, which means that the points $$A(1, 2, -1)$$, $$B(0, 1, 5)$$, $$C(-1, 2, 1)$$, and $$D(2, 1, 3)$$ all lie in the same plane. --- ### Do you need further clarification or details? Let me know! ### Related Questions: 1. What is the geometric significance of the scalar triple product? 2. How do you find the equation of a plane passing through three points? 3. Can coplanarity be determined using a determinant? 4. How do you test if vectors are linearly dependent? 5. How do you find the volume of a parallelepiped using vectors? ### Tip: When checking coplanarity of points in 3D, always consider the scalar triple product, as it’s a quick and reliable method.

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Math Problem Analysis

Mathematical Concepts

Vector Geometry
Coplanarity of Points
Scalar Triple Product
Cross Product

Formulas

Scalar Triple Product: u · (v × w) = 0 for coplanarity
Cross Product Determinant Formula

Theorems

Condition for Coplanarity: The scalar triple product of vectors formed by points must be zero.

Suitable Grade Level

Grade 10-12

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