Enrichment, Extension, and Application

Question 1

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Exercises

  1. Two \( 3 \)-dimensional vectors, \( \vec{u} \) and \( \vec{v} \), are placed tail to tail. Describe, geometrically, all three dimensional vectors \( \vec{w} \) that, when placed tail to tail with \( \vec{u} \) and \( \vec{v} \), form the same angle with both vectors simultaneously.
  2. (Note: These results may be used in later problems) Prove the following statements:
    Statement 1: If point \( P \) divides the segment \( AB \) in the ratio \( a : b \), then for any point \( O \), we have \[ \overrightarrow{OP} = \dfrac{b}{a + b}\overrightarrow{OA} + \dfrac{a}{a + b}\overrightarrow{OB} \]
    Statement 2: If, for a point \( O \), we have \( \overrightarrow{OP} = b\overrightarrow{OA} + a\overrightarrow{OB} \) with \( a + b = 1 \), then \( A\), \(P \), and \( B \) are collinear and \( P \) divides the segment \( AB \) in the ratio \( a : b \).
  3. In a quadrilateral \( ABCD \), \( T \) is the midpoint of the side \( AB \), \( U \) is the midpoint of the side \( CD \), \( L \) is the midpoint of the diagonal \( AC \), and \( M \) is the midpoint of the diagonal \( BD \). Denote \( \overrightarrow{AB} = \vec{a}\), \(\overrightarrow{BC} = \vec{b}\), and \(\overrightarrow{CD} = \vec{c} \).
    1. Show that \( \overrightarrow{AD} + \overrightarrow{BC} = 2\overrightarrow{TU} \).
    2. Show that \( \overrightarrow{AD} + \overrightarrow{CB} = 2\overrightarrow{LM} \).
    3. If \( ABCD \) is a trapezoid, with parallel bases \( BC = 5 \) and \( AD = 9 \), find
      1. the length of the line segment connecting the midpoints of the non-parallel sides; and
      2. the length of the line segment connecting the midpoints of the diagonals.
  4. In \( \triangle{ABC} \), \( AM\), \(BN \), and \( CP \) are medians. Prove that \( \overrightarrow{AM} + \overrightarrow{BN} + \overrightarrow{CP} = \vec{0} \).
  5. In \( \triangle{ABC} \), the altitudes from \( B \) and \( C \) meet at point \( H \). Denote \( \overrightarrow{AB} = \vec{b}\), \(\overrightarrow{AC} = \vec{c} \), and \( \overrightarrow{AH} = \vec{h} \).
    1. Express \( \overrightarrow{BH} \) and \( \overrightarrow{CH} \) in terms of \( \vec{b}\), \(\vec{c} \), and \( \vec{h} \).
      1. Use the dot product to express the fact that \( \overrightarrow{BH} \) and \( \overrightarrow{CH} \) are altitudes.
      2. Use the equations in part i) to show that \( AH \) is also an altitude, and hence that the altitudes in a triangle are concurrent.
  6. Definition: A vector \( \vec{u} \) is said to be a linear combination of some (other) vectors \( \vec{a}, \vec{b}, \dots \) if \( \vec{u} \) can be written as \[ \vec{u} = k_1\vec{a} + k_2\vec{b} + \cdots \] where \( k_1, k_2, \dots \) are constants.

    A non-zero vector situated on a line is said to be a base for the line, because any other vector on the line can be expressed uniquely in terms of the given vector. In higher dimensions, two non-collinear vectors in a plane form a base for the plane, because any vector in the plane can be expressed uniquely as a linear combination of those two vectors; similarly, three non-coplanar vectors form a base for the (three-dimensional) space.

    1. Regular octagon with vertices A,B,C,D,E,F,G,H

      In the diagram, \( ABCDEFGH \) is a regular octagon (recall: a regular octagon has all sides equal length and all angles equal measure), as shown below. Let \( \overrightarrow{AB} = \vec{a} \) and \( \overrightarrow{AH} = \vec{b} \). Since these vectors are not collinear, they form a basis for the plane, and all other vectors in the plane can be expressed as linear combinations of \( \vec{a} \) and \( \vec{b} \). Express the vectors along the other sides of the octagon ( \( \overrightarrow{BC}, \overrightarrow{CD}, \overrightarrow{DE}, \dots \)) as a sum of scalar multiples of \( \vec{a} \) and \( \vec{b} \).

    2. Cube with vertices A,B,C,D,O,E,F,G; vectors OB, OA, OC drawn

      Consider the cube \( ABCDEFGO \) shown. The vectors \( \vec{i} = \overrightarrow{OE}, \vec{j} = \overrightarrow{OG} \), and \( \vec{k} = \overrightarrow{OD} \) form a basis for the space as they are non-coplanar. Therefore, all vectors (in three-dimensional space) can be expressed uniquely as linear combinations of \( \vec{i}\), \(\vec{j} \), and \( \vec{k} \).

      1. Express the vectors \( \vec{a} = \overrightarrow{OA}\), \(\vec{b} = \overrightarrow{OB} \), and \( \vec{c} = \overrightarrow{OC} \) as linear combinations of \( \vec{i}\), \(\vec{j} \), and \( \vec{k} \).
      2. Observe that no pair of the vectors \( \vec{a}, \vec{b}, \vec{c} \) are collinear. Prove that they are not coplanar by showing that \( \vec{b} \) cannot be written as a linear combination of \( \vec{a} \) and \( \vec{c} \).
      3. As a result of part b), \( \vec{a}, \vec{b}, \vec{c} \) also form a basis for the space. Express \( \vec{i}\), \(\vec{j} \), and \( \vec{k} \) as linear combinations of \( \vec{a}\), \(\vec{b}\), and  \(\vec{c} \).
      4. Express the vector \( \overrightarrow{AG} \) as a linear combination of \( \vec{i}\), \(\vec{j} \), and \( \vec{k} \)
      5. Express the vector \( \overrightarrow{AG} \) as a linear combination of \( \vec{a}\), \(\vec{b} \), and \( \vec{c} \)
  7. In the figure below, \( ABCD \) and \( DEFG \) are parallelograms. Point \( D \) divides the segment \( AG \) in the ratio \( 1 : 2 \). Point \( E \) is the midpoint of the segment \( DC \). If \( BD \) is perpendicular to \( DF \), and \( \cos{\left( \angle{ADC} \right)} = 0.3 \), find the ratio \( AD : DE \).
    Image of figure described in question
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