# Calculus II/Test 4

If we follow the previous course, Test 4 might include material on the power series.

• All the examples in Chapter 9.1 might be on the test (pp634-38)
• wikipedia:Dot product wikipedia:Cross product
• Vectors let's look at rotations.--Guy vandegrift (discusscontribs) 16:56, 5 April 2017 (UTC)
• These examples from Chapter 9 look like good candidates for Test 4:
• Example 1 p664: Find the equation of a line (in x,y,z coordinates) in a given direction through a given point.
• Example 2 p665: Find the equation of a line (in x,y,z coordinates) through two points.
• Example 4 p667: Find the equation of a plane perpendicular to a given direction, passing through a given point.
• Polar coordinates Appendix H page A55
• Examples 2-5 are easy and at least one of these will be on the test.
• Example 6 is moderately difficult and will be on the test in one form or another.
• The whole class stopped at A58 and nothing on or after Example 7 will be covered.

### Example 8 Challenge question "extra credit' and not a lot of it.

See p669 of textbook: Find the distance from a point ${\displaystyle {\vec {r}}_{1}=\langle x_{1},y_{1},z_{1}\rangle }$  to given a plane, if the plane is defined as follows:

${\displaystyle {\vec {r}}_{0}=\langle x_{0},y_{0},z_{0}\rangle }$  is some point on the plane (i.e., three given numbers).
${\displaystyle {\vec {n}}=\langle a,b,c\rangle }$  is normal to the plane (again, three given numbers).

Now draw the point and the plane in from a specific angle in which the given point and the normal lie in the plane of the paper (board), and use facts about dot-product. (I never bothered to define comp and proj as described on 652; I was familiar with comp but never heard of proj, but it is an easy concept to grasp if you understand this problem).

Rule: If, ${\displaystyle {\hat {n}}={\vec {n}}/n}$  is a unit normal to the plane, then the magnitude of ${\displaystyle {\vec {R}}\cdot {\hat {n}}}$  is the distance from the point ${\displaystyle {\vec {r}}=\langle x_{1},y_{1},z_{1}\rangle }$  to the plane. In the figure:
1. ${\displaystyle {\mathcal {A}}={\vec {R}}={\vec {r}}_{1}-{\vec {r}}_{0}}$  is any vector connecting the point to somewhere on the plane (the calculated distance should not depend on ${\displaystyle {\vec {r}}_{0}}$  ).
2. The unit normal is given by: ${\displaystyle {\hat {n}}={\frac {\langle a,b,c\rangle }{\sqrt {a^{2}+b^{2}+c^{2}}}}}$
3. ${\displaystyle {\vec {R}}\cdot {\hat {n}}=\langle x_{1}-x_{0},y_{1}-y_{0},z_{1}-x_{0}\rangle \cdot {\frac {\langle a,b,c\rangle }{\sqrt {a^{2}+b^{2}+c^{2}}}}}$ ${\displaystyle {\frac {a(x_{1}-x_{0})+b(y_{1}-y_{0})+c(z_{1}-x_{0})}{\sqrt {a^{2}+b^{2}+c^{2}}}}}$

${\displaystyle {\frac {ax_{1}+by_{1}+cz_{1}-\{ax_{0}+by_{0}+cz_{0}\}}{\sqrt {a^{2}+b^{2}+c^{2}}}}}$

If we view the expression in curly brackets ${\displaystyle \{ax_{0}+by_{0}+cz_{0}\}=-d}$  as some constant, and drop the subscript "1", then we have the formula for a plane normal to ${\displaystyle {\hat {n}}}$ .

${\displaystyle ax+by+cz+d=0}$

Note that the distance, ${\displaystyle |{\vec {R}}\cdot {\hat {n}}|=0}$  if ${\displaystyle ax_{1}+by_{1}+cz_{1}+d=0}$  because the point is already on the plane.