MATH 2D Study Guide - Midterm Guide: Paraboloid

(a) Find a vector w with magnitude 20 that is parallel to v =<3, 4, 2>

b.) give the angle between vectors a=3i-j+5k and b=i+2j (Give anexact expression for your answer and then approximate to thenearest degree.)

Question1.

a)

Let L be the line containing thepoint P(2,-5,-1) and perpendicular to theplane 2x1+3x2+2x3=-3.

Find a vector equationfor L.

My answer: <2,-5,-1>+ t<2,3,2> (correct)

b)

At what pointdoes

Lintersect the yz-plane?

Question2:

x1 = -6 t - 4

x2 = - t+ 2

x3 = 5t - 6

and the second set ofparametric equations

x1 = 6t

x2 = ?

x3 = ?

describe the same line inR3. Find x2 and x3.

Question3:

It is one of 3 options foreach one: Parallel, perpindicular, or neither. Please explainwhy.

http://imgur.com/UbDuH8Z

Question4:

Find the point ofintersection of the two lines

a)

L1: x=1+t, y=-10-2t, z=-5-2t,

L2: x=7-4s, y=-10+2s, z=-7+3s.

Note that in general, twolines in 3D do not intersect. These two lines are carefully chosenso that they do

b)

Find the point where theline

x=1+2t,y=-4+t,z=-2-2t

intersects theplane

4x-y+3z=3.

Thanks. I have a lot ofpoints for this and will give it to someone that can helpme.

1.Write a function encipher(s, n) that takes as inputs an arbitrary string s and a non-negative integer n between 0 and 25, and that returns a new string in which the letters in s have been "rotated" by n characters forward in the alphabet, wrapping around as needed. For example:

>>> encipher('hello', 1) result: 'ifmmp'

>>> encipher('hello', 2) result: 'jgnnq'

>>> encipher('hello', 4) result: 'lipps'

Upper-case letters should be "rotated" to upper-case letters, even if you need to wrap around. For example:

>>> encipher('XYZ', 3)

result: 'ABC'

Lower-case letters should be "rotated" to lower-case letters:

>>> encipher('xyz', 3)

result: 'abc'

Non-alphabetic characters should be left unchanged:

>>> encipher('#caesar!', 2)

result: '#ecguct!'

Hints/reminders:

You can use the built-in functions ord and chr convert from single-character strings to integers and back:

>>> ord('a')

result: 97

>>> chr(97)

result: 'a'

You can use the following test to determine if a character is between 'a' and 'z' in the alphabet:

if 'a' <= c <= 'z':

A similar test will work for upper-case letters.

We recommend writing a helper function rot(c, n) that rotates a single character c forward by n spots in the alphabet. We have given you a template for this helper function in ps3pr3.py that checks to ensure that c is a single-character string. We wrote rot13(c) in lecture; rot(c, n) will be very close to rot13(c)! You can test your rot(c, n) as follows:

>>> rot('a', 1)

result: 'b'

>>> rot('y', 2)

result: 'a'

>>> rot('A', 3)

result: 'D'

>>> rot('Y', 3)

result: 'B'

>>> rot('!', 4)

result: '!'

Once you have rot(c, n), you can write a recursive encipher function.

Once you think you have everything working, here are three more examples to try:

>>> encipher('xyza', 1)

result: 'yzab'

>>> encipher('Z A', 2)

result: 'B C'

>>> encipher('Caesar cipher? I prefer Caesar salad.', 25)

result: 'Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.'

2.Write a function decipher(s) that takes as input an arbitrary string s that has already been enciphered by having its characters "rotated" by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s. For example:

>>> decipher('Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.')

result: 'Caesar cipher? I prefer Caesar salad.'

Here are two more examples:

>>> decipher('Hu lkbjhapvu pz doha ylthpuz hmaly dl mvynla lclyfaopun dl ohcl slhyulk.')

result: 'An education is what remains after we forget everything we have learned.'

>>> decipher('python')

result: 'eniwdc'

------------------------------------

def rot(c, n):
    """ your docstring goes here """
    # check to ensure that c is a single character
    assert(type(c) == str and len(c) == 1)

def letter_prob(c):
    """ if c is the space character (' ') or an alphabetic character,
        returns c's monogram probability (for English);
        returns 1.0 for any other character.
        adapted from:
       
    """
    # check to ensure that c is a single character   
    assert(type(c) == str and len(c) == 1)

    if c == ' ': return 0.1904
    if c == 'e' or c == 'E': return 0.1017
    if c == 't' or c == 'T': return 0.0737
    if c == 'a' or c == 'A': return 0.0661
    if c == 'o' or c == 'O': return 0.0610
    if c == 'i' or c == 'I': return 0.0562
    if c == 'n' or c == 'N': return 0.0557
    if c == 'h' or c == 'H': return 0.0542
    if c == 's' or c == 'S': return 0.0508
    if c == 'r' or c == 'R': return 0.0458
    if c == 'd' or c == 'D': return 0.0369
    if c == 'l' or c == 'L': return 0.0325
    if c == 'u' or c == 'U': return 0.0228
    if c == 'm' or c == 'M': return 0.0205
    if c == 'c' or c == 'C': return 0.0192
    if c == 'w' or c == 'W': return 0.0190
    if c == 'f' or c == 'F': return 0.0175
    if c == 'y' or c == 'Y': return 0.0165
    if c == 'g' or c == 'G': return 0.0161
    if c == 'p' or c == 'P': return 0.0131
    if c == 'b' or c == 'B': return 0.0115
    if c == 'v' or c == 'V': return 0.0088
    if c == 'k' or c == 'K': return 0.0066
    if c == 'x' or c == 'X': return 0.0014
    if c == 'j' or c == 'J': return 0.0008
    if c == 'q' or c == 'Q': return 0.0008
    if c == 'z' or c == 'Z': return 0.0005
    return 1.0