18.02, Problem set III. Due Feb 18, 12:45 in 2-106

If you haven't done so already, remember to join the 1802 list. There is an in-class test Friday, Feb 19 - which is why this problem set is due on Thursday, Feb 18. Material in the lecture on Thursday Feb 18 will not be included in the test - so it will be based on the first 6 lectures, sections 10.4, 12.1-12.6 of Edwards and Penney and parts L and K of the supplementary notes.

Problems are from the text (EP=Edwards and Penney) or from the supplementary notes (SN).

Hand in underlined problems from Part I and all of Part II. Note that the solutions to Part I problems are generally available in the notes, Section S. Part II is marked more critically with points as indicated. These points will accumulate and finally constitute 30% of the possible total.

Part I (1 pt each for correct answers)

Lec 5 (Thurs Feb 11): Read EP Sect 12.4, 10.4 to p. 591, 12.5 to p. 747. Problems:EP p. 742 nos. 3, 7, 22, 33. EP p. 594, nos. 4, 12, 15. EP p. 755 nos. 3, 4, 13, 31, 39, 40.

Lec 6 (Fri Feb 12): Read EP Sect 12.6, SN K. Problems SN p. K.2, nos. 1, 2, 3.

Part II

Problem 1: (3 pts) Given that $\bf A=2\bf i+\bf j+3\bf k$ and $\bf B=3\bf i-2\bf j-\bf k$ find all vectors of length $4$ which are perpendicular to both $\bf A$ and $\bf B.$

Solution: The cross product is ${\bf A}\times{\bf B}=(-1+6){\bf i}-(-2-9){\bf j}+(-4-3){\bf k}=5{\bf i}+11{\bf j}-7{\bf k}.$ This has length $\sqrt{195}$ so the two vectors of length $4$ perpendicular to both ${\bf A}$ and ${\bf B}$ are

$$\pm\frac{2}{\sqrt{195}}(5{\bf i}+11{\bf j}-7{\bf k}).$$

Problem 2: (2+2 pts) Let $O$ be the origin, $P=(2,0,1)$ and let $L$ be the line through the origin parallel to $2\bf i-\bf j+2\bf k.$

a) Express the vector $\bf {OP}$ as the sum of a vector parallel to $L$ and a vector perpendicular to $L.$

b) Calculate the distance from $P$ to $L.$

Solution:

a) ${\bf OP}=2{\bf i}+{\bf k}$ and ${\bf A}=2\bf i-\bf j+2\bf k$ is parallel to the line. Since ${\bf A}$ has length $3$ the vector $\frac19({\bf OP}\cdot{\bf A}){\bf A}=\frac23{\bf A}$ has the same dot product with ${\bf A}$ as ${\bf OP}$ has (namely $6.)$ Thus

$${\bf OP}=\frac23(2\bf i-\bf j+2\bf k)+\frac13(2{\bf i}+2{\bf j}-{\bf k})$$

is the (only) decomposition into a part parallel to the line and a part perpendicular to it.

b) A general point on the line is $\frac13(2{\bf i}+2{\bf j}-{\bf k})+t{\bf A}$ (since it is ${\bf OP}$ plus some multiple of ${\bf A}).$ The length of this is greater than or equal to $\vert\frac13(2{\bf i}+2{\bf j}-{\bf k})\vert=1$ which is therefore the distance from the origin to the line.

Problem 3: (2+2 pts)

Consider the system $x_1+x_2=2cx_1,$ $x_1+2x_2+x_3=2cx_2,$ $x_2+x_3=2cx_3.$

a) For what values of the constant $c$ will there be a non-trivial solution?

b) Let $c=0.$ Find a non-trivial solution by writing the three equations in vector form as ${\bf A}\cdot {\bf x}=0,$ ${\bf B}\cdot {\bf x}=0,$ and ${\bf C}\cdot {\bf x}=0,$ then using vector analysis to find a non-zero vector ${\bf x}$ which is orthogonal to all three vectors ${\bf A},$ ${\bf B}$ and ${\bf C}.$

Solution:

a) The determinant of the matrix $\left[\begin{matrix}(1-2c)&1&0\\ 1&(2-2c)&1\\ 0&1&(1-2c)\end{matrix}\right]$ is

$$2(1-c)(1-2c)^2-2(1-2c)=2(1-2c)(2c^2-3c+1-1)=2c(1-2c)(2c-3).$$

Thus the determinant vanishes, and hence there is a non-trivial solution, if $c=0,$ $c=\frac12$ or $c=\frac32.$

b) If $c=0$ then the equations are ${\bf A}\cdot {\bf x}=0,$ ${\bf B}\cdot {\bf x}=0,$ and ${\bf C}\cdot {\bf x}=0,$ if ${\bf A}={\bf i}+{\bf j},$ ${\bf B}={\bf i}+2{\bf j}+{\bf k}$ and ${\bf C}={\bf j}+{\bf k}.$ The cross product of ${\bf A}$ and ${\bf C}$ is ${\bf i}-{\bf j}+{\bf k}$ - which is orthogonal to all three. Thus a non-trivial solution is $x_1=1,$ $x_2=-1$ and $x_3=1.$

Problem 4: (8 pts)

Does a pitched ``curve ball'' in baseball really curve? Work through project 12.5 in EP. There a several questions to be answered; in your answer label then (a), (b) and so on, in order. Write up the last two in decent English.