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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 1.4 Week 1 Activity

Subsection 1.4.1 Vector Arithmetic

Activity 1.4.1.

For the vectors \(u = (4,5)\) and \(v = (-1, -3)\text{,}\) calculate the operations. Draw the first three.
  1. \begin{equation*} u + v \end{equation*}
  2. \begin{equation*} u - v \end{equation*}
  3. \begin{equation*} \frac{1}{4} u - \frac{3}{5} v \end{equation*}
  4. \begin{equation*} |u| \end{equation*}
  5. \begin{equation*} |u - v| \end{equation*}
  6. \begin{equation*} u \cdot v \end{equation*}
  7. The angle between \(u\) and \(v\text{.}\)
Solution.
  1. \begin{equation*} u + v = (4 + (-1)1, 5 + (-3)) = (3,2) \end{equation*}
  2. \begin{equation*} u - v = (4 - (-1), 5 - (-3)) = (5, 8) \end{equation*}
  3. \begin{equation*} \frac{1}{4} u - \frac{3}{5} v = \left(1, \frac{5}{4} \right) - \left( \frac{-3}{5}, \frac{-9}{5} \right) = \left( \frac{8}{5}, \frac{61}{20} \right) \end{equation*}
  4. \begin{equation*} |u| = \sqrt{4^2 + 5^2} = \sqrt{41} \end{equation*}
  5. \begin{equation*} |u - v| \sqrt{5^2 + 8^2} = \sqrt{89} \end{equation*}
  6. \begin{equation*} u \cdot v = (4)(-1) + (5)(-3) = -19 \end{equation*}
  7. The angle between \(u\) and \(v\text{.}\)
    \begin{equation*} \cos \theta = \frac{u \cdot v}{|u||v|} = \frac{-19}{\sqrt{41}\sqrt{13}}\text{.} \end{equation*}
    Inverse cosine gives \(\theta \doteq 2.54\text{.}\)
Here is the diagram of the first three calculations.
Figure 1.4.1. Vector Operatiors Activity 1

Subsection 1.4.2 Equations of Planes and Hyperplanes

Activity 1.4.2.

Write the equation of the plane with normal \((-1,-1,0)\) if \((0,1,3)\) is a point on the plane.
Solution.
The normal determines the coefficients of the equation of the plane. I put those coefficients into the general form, with the constant \(d\) undetermined.
\begin{equation*} (-1)x + (-1)y + 0 z = d \end{equation*}
I can calculate the constant using the point. I put the coordinates of the point into the left side of the equation and calculate \(d\text{.}\)
\begin{equation*} d = (-1)(0) + (-1)(1) + (0)(3) = -1 \end{equation*}
Therefore, the equation of the plane is
\begin{equation*} - x - y = -1 \text{.} \end{equation*}

Activity 1.4.3.

Write the equation of the plane with normal \((2,0,3)\) if \((-4,-1,2)\) is a point on the plane.
Solution.
The normal determines the coefficients of the equation of the plane. I put those coefficients into the general form, with the constant \(d\) undetermined.
\begin{equation*} 2x + 0y + 3z = d \end{equation*}
I can calculate the constant using the point. I put the coordinates of the point into the left side of the equation and calculate \(d\text{.}\)
\begin{equation*} d = 2(-4) + 0(-1) + 3(2) = -2 \end{equation*}
Therefore, the equation of the plane is
\begin{equation*} 2x + 3z = -2 \text{.} \end{equation*}

Activity 1.4.4.

Write the equation of the plane with local directions \((-1,-1,1)\) and \((-2,1,0)\) if \((2,1,1)\) is a point on the plane.
Solution.
The cross product of the local directions gives the normal to the plane.
\begin{align*} (-1,-1,1) \times (-2,1,0) \amp = \left( (-1)(0) - (1)(1), (1)(-2) - (-1)(0), (-1)(1) - (-1)(-2) \right) \\ \amp = (-1, -2, -3) \end{align*}
This is the normal. The normal determines the coefficients of the equation of the plane. I put those coefficients into the general form, with the constant \(d\) undetermined.
\begin{equation*} -x - 2y - 3z = d \end{equation*}
I can calculate the constant using the point. I put the coordinates of the point into the left side of the equation and calculate \(d\text{.}\)
\begin{equation*} d = -(1)(2) - (2)(1) - 3(1) = -7 \end{equation*}
Therefore, the equation of the plane is
\begin{equation*} -x - 2y - 3z = -7 \text{.} \end{equation*}

Activity 1.4.5.

Write the equation of the plane with local directions \((0,1,1)\) and \((-2,1,1)\) if \((-1,0,3)\) is a point on the plane.
Solution.
The cross product of the local directions gives the normal to the plane.
\begin{align*} (0,1,1) \times (-2,1,1) \amp = \left( (1)(1) - (1)(1), (1)(-2) - (0)(1), (0)(1) - (1)(-2) \right) \\ \amp = (0,-2,2) \end{align*}
This is the normal. The normal determines the coefficients of the equation of the plane. I put those coefficients into the general form, with the constant \(d\) undetermined.
\begin{equation*} 0x - 2y + 2z = d \end{equation*}
I can calculate the constant using the point. I put the coordinates of the point into the left side of the equation and calculate \(d\text{.}\)
\begin{equation*} d = 0(-1) - 2(0) + 2(-3) = -6 \end{equation*}
Therefore, the equation of the plane is
\begin{equation*} -2y + 2z = -6 \text{.} \end{equation*}

Activity 1.4.6.

Write the equation of the plane with points \(p = (0,1,1)\text{,}\) \(q = (1,0,0)\) and \(r = (-2,1,1)\text{.}\)
Solution.
The local directions are given by the differences of the point: \((q-p)\) and \((r-p)\text{.}\) I calculate those two local directions.
\begin{gather*} q-p = (1,0,0) - (0,1,1) = (1,-1,-1)\\ r-p = (-2,1,1) - (0,1,1) - (-2,0,0) \end{gather*}
These are the two local directions. The cross product of the local directions gives the normal to the plane.
\begin{align*} (1,-1,-1) \times (-2,0,0) \amp = \left( (-1)(0) - (-1)(0), (-1)(-2) - (1)(0), (1)(0) - (-1)(-2) \right) \\ \amp = (0,2,-2) \end{align*}
This is the normal. The normal determines the coefficients of the equation of the plane. I put those coefficients into the general form, with the constant \(d\) undetermined.
\begin{equation*} 0x + 2y - 2z = d \end{equation*}
I can calculate the constant using the point (I can use any of the three points — all will determine the same constant). I put the coordinates of the point into the left side of the equation and calculate \(d\text{.}\)
\begin{equation*} d = (0)(0) + 2 (1) - 2(1) = 0 \end{equation*}
Therefore, the equation of the plane is
\begin{equation*} 2y - 2z = 0 \text{.} \end{equation*}

Activity 1.4.7.

Write the equation of the plane with points \(p = (-3,-1,0)\text{,}\) \(q = (-2,1,4)\) and \(r = (0,5,0)\text{.}\)
Solution.
The local directions are given by the difference of the points: \((q-p)\) and \((r-p)\text{.}\) I calculate those two local directions.
\begin{gather*} q-p = (-2,1,4) - (-3, -1, 0) = (1,2,4)\\ r-p = (0,5,0) - (-3,-1,0) = (3,6,0) \end{gather*}
These are the two local directions. The cross product of the local directions gives the normal to the plane.
\begin{align*} (1,2,4) \times (3,6,0) \amp = \left( (2)(0) - (-4)(6), (-4)(3) - (1)(0), (1)(6) - (2)(3) \right) \\ \amp = (-24,12,0) \end{align*}
This is the normal. The normal determines the coefficients of the equation of the plane. I put those coefficients into the general form, with the constant \(d\) undetermined.
\begin{equation*} -24 x + 12 y + 0z = d \end{equation*}
I can calculate the constant using the point (I can use any of the three points — all will determine the same constant). I put the coordinates of the point into the left side of the equation and calculate \(d\text{.}\)
\begin{equation*} d = -24(0) + 12(5) + 0 z = 60 \end{equation*}
Therefore, the equation of the plane is
\begin{equation*} -24 x + 12 y = 60 \text{.} \end{equation*}

Subsection 1.4.3 Conceptual Review Questions

  • How does vector arithmetic differ from number arithmetic?
  • What is a cross product and what does it mean?
  • What is a dot product and what does it mean?
  • What is a normal vector to a plane? Why are planes determined by normals?
  • Why do cross products help calculate normals to planes in three dimensions?
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