Problem On Finding A Vector Perpendicular To A Triangle

Problem on finding a vector perpendicular to a triangle

$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$ Consider the triangle in three-space given by $\langle a, 0, 0\rangle, \langle 0, b, 0 \rangle, \langle 0, 0, c \rangle.$ Find a vector that is perpendicular to the triangle and has length equal to the area of the triangle.

• Solution

  Recall that
  • Cross product gives a perpendicular vector
  The cross product of two vectors is perpendicular to both. Hence, if we can find two vectors within the triangle, a cross product will get us close to the vector we seek. We consider the vectors going from $\langle a, 0, 0\rangle$ to $\langle 0, b, 0\rangle$ or $\langle 0, 0, c\rangle.$ Recall that
  • Vector-scalar multiplication scales the length of the vector
  The vector going from $\bfx$ to $\bfy$ is $\bfy - \bfx$. The vectors from $\langle a, 0, 0\rangle$ to $\langle 0, b, 0\rangle$ or $\langle 0, 0, c\rangle$ are then $\langle -a, b, 0\rangle$ and $\langle -a, 0, c\rangle$: Hence a vector perpendicular to the triangle is $$\langle -a, b, 0 \rangle \times \langle -a, 0, c \rangle.$$
  • Cross product definition
  • Determinant of a 3x3 matrix
  Recalling the definition of the cross product, we have \begin{align}\langle -a, b, 0 \rangle \times \langle -a, 0, c \rangle &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix}\\ \ \\ &= \begin{vmatrix}b & 0 \\ 0 & c \end{vmatrix} \ \mathbf{i} - \begin{vmatrix}-a & 0 \\ -a & c \end{vmatrix} \ \mathbf{j} + \begin{vmatrix}-a & b \\ -a & 0 \end{vmatrix} \ \mathbf{k} \end{align}
  • Determinant of a 2x2 matrix
  Evaluating the $2 \times 2$ determinants, we get the perpendicular vector $$\langle -a, b, 0 \rangle \times \langle -a, 0, c \rangle = bc \ \bfi + ac \ \bfj + ab \ \bfk.$$ We need to relate this vector to the length of the area of the triangle. So, we seek the relationship of a cross product to the area of some shape. Recall that
  • Cross product and area of parallelograms
  The length $\left | \bfx \times \bfy \right|$ equals the area as the parallelegram spanned by $\bfx$ and $\bfy.$ We note that the parallelegram spanned by two vectors has twice the area as the triangle spanned by them: Hence the perpendicular vector whose length equals the area of the triangle is: $$\frac{\langle -a, b, 0 \rangle \times \langle -a, 0, c\rangle}{2} =\left \langle \frac{bc}{2}, \frac{ac}{2}, \frac{ab}{2} \right \rangle.$$

Related topics

• Multivariable calculus (147 problems)
  • Vectors (55 problems)
    • Vector subtraction (20 problems)
    • Vector scalar multiplication (13 problems)
      • $c \mathbf{x}$ is the vector obtained by multiplying the length of $\bfx$ by $c$. (6 problems)
  • Determinants (23 problems)
    • Determinant of a 3x3 matrix (15 problems)
    • Determinant of a 2x2 matrix (16 problems)
  • Cross product (17 problems)
    • The cross product is defined by $\mathbf{x} \times \mathbf{y} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{vmatrix} $. (12 problems)
    • The cross product of two vectors is a vector perpendicular to both. (7 problems)
    • The area of the parallelogram spanned by the vectors $\mathbf{x}$ and $\mathbf{y}$ is $| \mathbf{x} \times \mathbf{y}|$. (3 problems)

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