Planes and the Cross Product

Here are the questions we considered in today’s class. See if you can answer these (with algebraic solutions) for tomorrow’s lesson. GeoGebra will be useful to check you answers, and to give you some insight into the question if you get stuck.

  1. Find a vector normal to both
    \[\vec{a}=\begin{bmatrix}1\\2\\-2\end{bmatrix} \text{ and }\vec{b}=\begin{bmatrix}3\\4\\1\end{bmatrix}.\]
  2. a) Find a vector normal to the plane \[\vec{r}=\begin{bmatrix}1\\4\\-2\end{bmatrix} +\lambda \begin{bmatrix}1\\1\\-1\end{bmatrix} + \mu \begin{bmatrix}-3\\1\\2\end{bmatrix}\] b) Using your answer to part a), can you find the distance of the point \(A(1, 1, 1)\) to the given plane?

Planes

Try to complete the following questions for our lesson on Monday.

  1. Find a vector equation of the plane passing through \(A(1,2,3)\), \(B(3,1,-2)\), and \(C(4,4,4)\).
  2. Find a vector equation of the plane with Cartesian equation \(3x+2y-z=1\).
  3. Find a vector equation of the plane containing the line \[\vec{r}=\begin{bmatrix}-1\\-1\\4\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}\] and passing through the point \(A(6,-3,2)\).

Lines in 3D

Try to complete the following questions for Monday’s lesson.

  1. Find the distance of the point  \(A(-1,1,2)\) to the line \[\displaystyle{\vec{r}=\begin{bmatrix}1\\-2\\3\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}}\]
  2. The lines below intersect at a point \(A\). Use an algebraic method to find the coordinates of  \(A\), then verify your answer using GeoGebra.
    \[L_1: \vec{r}=\begin{bmatrix}-1\\-1\\4\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}\] \[L_2: \vec{r}=\begin{bmatrix}-2\\0\\7\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}\]
  3. In 2D, any two non-parallel lines will have a point of intersection. In 3D, two lines can be non-parallel and have no point of intersection; such lines are called skew lines.
    Show that \(L_1\) from question 2 and \(L_3\) below are skew lines.
    \[L_3: \vec{r}=\begin{bmatrix}-2\\5\\12\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}\]

Vector Equations of Lines 3

Find the distance of the point \(A(-3,1)\) to the line
\[\displaystyle{\vec{r}=\begin{bmatrix}1\\2\end{bmatrix}+\lambda \begin{bmatrix}-2\\3\end{bmatrix}}\]

using the method outlined below.

  1. Find an expression for an arbitrary point \(D\) on the given line.
  2. Using the expression you’ve produced, find a general expression for the vector \(\overrightarrow{AD}\).
  3. Where \(\vec{b}\) is the direction vector of the given line, use the dot product \(\overrightarrow{AD}\cdot\vec{b}\) to find the coordinates of the point on the line closest to \(A\). Hence, find the distance from \(A\) to the line.

Vector Equations of Lines 2

Complete the following question for our lesson tomorrow. In the first instance, use the Cartesian equation to find your solution, then see of you can find a vector-based solution to the same problem.

Find the distance of the point \(A(1,3)\) to the line
\[\displaystyle{\vec{r}=\begin{bmatrix}4\\2\end{bmatrix}+\lambda \begin{bmatrix}4\\1\end{bmatrix}}\]