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Give a Cartesian for planes through given points spanned by given vectors

Consider the vectors

    \[ A = 2 \mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k}, \qquad B = \mathbf{j} + \mathbf{k}. \]

  1. Find a nonzero vector N perpendicular to both A and B.
  2. Find a Cartesian equation for the plane through (0,0,0) which is spanned by A and B.
  3. Find a Cartesian equation for the plane through (1,2,3) which is spanned by A and B.

  1. Since A and B are independent, we can take

        \[ N = A \times B = (3 - (-4), 0 - 2, 2 - 0) = (7,-2,2). \]

  2. From part (a) we have N = (7,-2,2) is perpendicular to both A and B, so a Cartesian equation for the plane is given by

        \[ 7x - 2y + 2z = d \]

    Further, since the point (0,0,0) is on the plane, we must have d = 0. Hence, the Cartesian equation for the plane is

        \[ 7x - 2y + 2z = 0. \]

  3. Again, we have a Cartesian equation for the plane given by

        \[ 7x - 2y + 2z = d. \]

    Since (1,2,3) is on the plane we must have

        \[ d = 7(1) - 2(2) + 2(3) \quad \implies \quad d = 9. \]

    Hence, the Cartesian equation for the plane is given by

        \[ 7x - 2y + 2z = 9. \]

Give a vector based proof of Heron’s formula for computing the area of a triangle

Let S denote the area of a triangle with sides of lengths a,b,c. Heron’s formula states that

    \[ S = \sqrt{s (s-a)(s-b)(s-c)}, \qquad \text{where} \qquad s = \frac{a+b+c}{2}. \]

We prove this formula using vectors via the following steps. Assume the triangle has vertices O,A, and B with

    \[ \lVert A \rVert = a, \qquad \lVert B \rVert = b, \qquad \lVert B - A \rVert = c. \]

  1. Using the identities

        \[ \lVert A \times B \rVert^2 = \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2, \qquad -2A \cdot B = \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \]

    prove the formula

        \[ 4S^2 = a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 = \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2). \]

  2. Simplify the formula in part (a) to obtain the formula

        \[ S^2 = \frac{1}{16}(a+b+c)(a+b-c)(c-a+b)(c+a-b), \]

    and use this to deduce Heron’s formula.


  1. Proof. We know the area of the triangle (in terms of the vectors A and B) is

        \[ S = \frac{1}{2} \lVert A \times B \rVert. \]

    Therefore,

        \begin{align*}  4S^2 &= \lVert A \times B \rVert^2 \\  &= \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2 \\  &= a^2 b^2 - \frac{1}{4} ( \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 ) \\  &= a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 \\  &= \frac{1}{4} (4a^2 b^2 - (c^2 - a^2 - b^2)^2) \\  &= \frac{1}{4} (2ab - (c^2 - a^2 - b^2))(2ab + (c^2 - a^2 - b^2)) \\  &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2). \qquad \blacksquare \end{align*}

  2. Proof. We simplify the formula in part (a),

        \begin{align*}  && 4S^2 &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2) \\  \implies && S^2 &= \frac{1}{16} ((a+b)^2 - c^2)(c^2 - (a-b)^2) \\  \implies && S^2 &= \frac{1}{16} (a+b+c)(a+b-c)(c-a+b)(c+a-b) \\  \implies && S^2 &= \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} \cdot \frac{c-a+b}{2} \cdot \frac{c+a-b}{2} \\[9pt]  \implies && S^2 &= s(s-c)(s-a)(s-b) \\  \implies && S &= \sqrt{s (s-a)(s-b)(s-c)}. \qquad \blacksquare \end{align*}

Prove a formula for the perpendicular distance between a point and a line

  1. Assume that B \neq C. Prove that the perpendicular distance from A to the line passing through the points B and C is given by the formula

        \[ \frac{\lVert (A - B) \times (C - B) \rVert}{\lVert B - C\rVert}. \]

  2. Compute the distance in the case

        \[ A = (1,-2,-5), \qquad B = (-1,1,1), \qquad C = (4,5,1). \]


  1. Proof. We know the area of the parallelogram determined by (A-B) and (C-B) is given by

        \[ \lVert (A-B) \times (C-B) \rVert. \]

    But this is exactly twice the area of the triangle with base C-B and height the perpendicular line from A to the line through B and C. Hence, h is the perpendicular distance from A to the line through B and C, and

        \[ \lVert (A-B) \times (C-B) \rVert = 2 \cdot \left( \frac{1}{2} h \lVert B-A \rVert \right) \quad \implies \quad h = \frac{\lVert (A-B) \times (C-B) \rVert}{\lVert B-C \rVert}. \qquad \blacksquare \]

  2. When A = (1,-2,-5), \ B = (-1,1,1), \ C = (4,5,1) we have

        \begin{align*}  \frac{\lVert (A-B) \times (C - B) \rVert}{\lVert B-C \rVert} &= \frac{\lVert (2,-3,-6) \times (5,4,0) \rVert}{\lVert (-5,-4,0) \rVert} \\[9pt]  &= \frac{\lVert (24,-30,23) \rVert}{\lVert (-5,-4,0) \rVert} \\[9pt]  &= \frac{\sqrt{2005}}{\sqrt{41}}.  \end{align*}

Prove a vector formula for the volume of a tetrahedron

  1. Consider a tetrahedron with vertices A,B,C,D. Prove that the volume of the tetrahedron is given by the formula

        \[ V = \frac{1}{6} |(B-A) \cdot (C -A) \times (D-A)|. \]

  2. Compute the volume in the case that

        \[ A = (1,1,1), \quad B = (0,0,2), \quad C = (0,3,0), \quad D = (4,0,0). \]


  1. Proof. We know the volume of a tetrahedron is given by \frac{1}{3} A_0 h (where A_0 denotes the altitude of the tetrahedron). We know (page 490 of Apostol) that the volume of the parallelepiped with base formed by vector A,B and height formed by vector C is given by A \times B \cdot C. In this case we have that the base of the tetrahedron is formed by the vectors (B-A) and (C-A), and the height is formed by the vector (D-A). Further, we know that the area of the base described by the vectors (B-A) and (C-A) is one half that of the parallelepiped whose base is given by vectors A and B (since the base of the parallelepiped described by vectors A and B is a rectangle, and the base of the tetrahedron is the triangle formed by cutting this rectangle along the diagonal). Therefore we have

        \begin{align*}  V &= \frac{1}{3} \left( \frac{1}{2} \right) (B -A)\cdot (C-A) \times (D - A) \\  &= \frac{1}{6} \big( (B-A) \cdot (C-A) \times (D-A) \big). \qquad \blacksquare \end{align*}

  2. Using the formula in part (a) with the given values of A,B,C,D we have

        \begin{align*}  V &= \frac{1}{6} \big( (-1,-1,1) \cdot (-1,2,-1) \times (3,-1,-1) \big) \\   &= \frac{1}{6} \big( (-1,-1,1) \cdot (-2-1, -3-1, 1-6) \big) \\  &= \frac{1}{6} \big( (-1,-1,1) \cdot (-3,-4,-5) \big) \\  &= \frac{1}{6} (3 + 4 - 5) \\  &= \frac{1}{3}. \end{align*}