Math Problem Statement

Given the geometrical points A(2, 1, -1), B(1, 3, -3), C(3, 1, 0), and D(7, -1, -6): Solve the problems involving scalar, vector, and mixed products to find the external angle at A, the median length BM, the area of triangle ABC, the height DH of the tetrahedron DABC, and the parallelogram area based on vectors a and b.

Solution

The task involves solving problems related to scalar, vector, and mixed products for the given geometrical figures and points in space.

Here’s a breakdown of the tasks:


Task 1: (Given points A(2,1,1)A(2, 1, -1), B(1,3,3)B(1, 3, -3), C(3,1,0)C(3, 1, 0), D(7,1,6)D(7, -1, -6))

  1. Find the external angle at vertex AA in triangle ABCABC:

    • This involves computing the scalar product between vectors AB\overrightarrow{AB} and AC\overrightarrow{AC}, normalizing them, and finding the angle using: cosθ=ABACABAC\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\|\overrightarrow{AB}\| \cdot \|\overrightarrow{AC}\|} The external angle is 180θ180^\circ - \theta.
  2. Find the length of the median BMBM in triangle ABCABC:

    • The median splits the opposite side into two equal halves. First, calculate the midpoint MM of ACAC using: M=(x1+x22,y1+y22,z1+z22)M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)
    • Then find the length of BMBM using the distance formula: BM=(xBxM)2+(yByM)2+(zBzM)2\|\overrightarrow{BM}\| = \sqrt{(x_B - x_M)^2 + (y_B - y_M)^2 + (z_B - z_M)^2}
  3. Find the area of triangle ABCABC:

    • Use the cross product of vectors AB\overrightarrow{AB} and AC\overrightarrow{AC}. The area is: Area=12AB×AC\text{Area} = \frac{1}{2} \|\overrightarrow{AB} \times \overrightarrow{AC}\|
  4. Find the height DHDH of tetrahedron DABCDABC:

    • The height is perpendicular from DD to the base plane ABCABC. Use the volume formula for a tetrahedron: V=16DA(AB×AC)V = \frac{1}{6} |\overrightarrow{DA} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})| Then, solve for the height DHDH using: V=13Base AreaHeightV = \frac{1}{3} \cdot \text{Base Area} \cdot \text{Height}

Task 2:

Calculate the area of a parallelogram formed by vectors: a=6p3q,b=3p+2q\vec{a} = 6\vec{p} - 3\vec{q}, \quad \vec{b} = 3\vec{p} + 2\vec{q} where p=3|\vec{p}| = \sqrt{3}, q=5|\vec{q}| = 5, and the angle between p\vec{p} and q\vec{q} is 2π3\frac{2\pi}{3}.

  • First, calculate a×b\vec{a} \times \vec{b} using vector cross product properties.
  • The area of the parallelogram is the magnitude of the cross product: Area=a×b\text{Area} = \|\vec{a} \times \vec{b}\|

Would you like detailed solutions and step-by-step calculations for any or all of these tasks?

Here are 5 related questions to explore further:

  1. How do we calculate the angle between two vectors in 3D space?
  2. What is the geometric interpretation of the cross product of two vectors?
  3. How can the volume of a tetrahedron be calculated using vector operations?
  4. How do the coefficients of vectors p\vec{p} and q\vec{q} affect the parallelogram's area?
  5. What is the difference between an external and internal angle in geometry?

Tip: Always verify vector directions when using cross and scalar products to ensure correct results!

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Math Problem Analysis

Mathematical Concepts

Scalar Product
Vector Product
Mixed Product
Geometry of Triangles and Tetrahedra

Formulas

cos θ = (u • v) / (||u|| ||v||)
Length of vector ||v|| = sqrt(vx^2 + vy^2 + vz^2)
Cross product ||u x v|| = ||u|| ||v|| sin(θ)
Area of triangle = 0.5 ||u x v||
Volume of tetrahedron = (1/6) |u • (v x w)|

Theorems

Law of Cosines in 3D
Geometric interpretation of cross product
Volume formula for a tetrahedron

Suitable Grade Level

Grades 10-12