# 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. Here the dotted line is the plane, and the normal is B, and r1 − r0 is A. The point r1 is tail vector A.

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

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

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

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

$ax+by+cz+d=0$

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