Question

Calculus III Questions:
1. Let Q = (1, -1, 2), P = (1, 3, -2) and N = <1, 2, 2>. Find the point of the intersection of the line through P in the direction of N, and the plane through Q perpendicular to N.

2. Find the distance between the indicated point and plane.
(a) (1, 1, 2) and 3x + y - 5z = 2
(b) (3, -2, 1) and the yz-plane

Answer 1

you got that one figured out :)

Answer 2

haha hello again. I was hoping maybe someone could do #1 again. if not, think you could help with #2?

Answer 3

1 is hopeless til you get the notation cleared up

2 is better

Answer 4

2 is in the cross product chapter, so I assume cross product use is required

Answer 5

the distance from a plane is defined perpendicular to it; so we need to use the normal vector from the plane, create a line equation from the point and see where it intersects the plane at; then measure the distance between the points

Answer 6

write me up the line equations

Answer 7

the normal for 2a would be <3, 1, -5>

Answer 8

good

Answer 9

parametric form:
x = 1 + 3t
y = 1 + t
z = 2 -5t

Answer 10

now use those values for xyz into the plane equation to find the point it intersects at

Answer 11

or rather the "t" value for the line

Answer 12

substitutiing the parametric forms into the equation of the plane I end up with 35t = 8, or t = 8/35

Answer 13

ah, so then I just plug it into the parametric form

Answer 14

3(1+3t) +(1+t) - 5(2-5t) = 2

(3+1-10)+ t(9+1+25) = 2


(-6)+ t(35) = 2

yep; t=8/35

Answer 15

yep; plug t into the parametric to get the point that is under the other one to measure to :)

Answer 16

plugging it in I get 59/35, 43/35, -30/35

Answer 17

im curious; if we take t <normal> and get the magnitude, will it be the same as the distance since we are anchored to 1,1,2 ?

Answer 18

i'm not sure.

Answer 19

t tells us how far to stretch the normal to get to any point from our anchored point right?

Answer 20

just checked the book, and it says the answer is: \[8/\sqrt{35}\]

Answer 21

for 2a

Answer 22

so we did something right, and something wrong

Answer 23

so the normal scaled by t should be the vector from point to point;

Answer 24

8/35 <3, 1, -5> = sqrt((24)^2+(8)^2+(40)^2)/35

http://www.wolframalpha.com/input/?i=sqrt%28%2824%29%5E2%2B%288%29%5E2%2B%2840%29%5E2%29%2F35

8/sqrt(35)

Answer 25

t is good, so after finding t is where it might have gone bad ...

Answer 26

oh alright, so in the future just multiply the t value times the normal?

Answer 27

IF the point that we are measureing is used to construct out line eqs; then yes

Answer 28

lets finish the long version tho :)

59/35, 43/35, -30/35 is out point on the plane right?

Answer 29

i believe so

Answer 30

(1,1,2) = (35/35, 35/35, 70/35)

( 35, 35, 70)/35
(-59, -43, 30)/35
----------------
(-16, -8,100)/35

yeah, i think something went bad in the mathing after finding "t"

Answer 31

so, note to selfs, magnitude t<n> :)

Answer 32

most likely yes haha

Answer 33

so i'm really doing t * ||normal||

Answer 34

rather than just t * normal

Answer 35

im not sure if t|n| is good; lets see, ive never tried

\[\frac{8}{35}|<3,1,-5>|=\frac{8}{35} \sqrt{9+1+25}=\frac{8}{35} \sqrt{35}\]
that works too

Answer 36

how exactly did you do it originally then? when you give the wolfram alpha answer

Answer 37

scaled n by t then magged it

Answer 38

so if my professor gives me a similar question on the test, I know how to find t easily, once I find t, what do i do exactly? do I multiply it by the magnitude of the normal?

Answer 39

\[|\frac{8}{35}<3,1,-5>|=|<\frac{24}{35},\frac{8}{35},-\frac{40}{35}>|\]

yes t|n| is good

Answer 40

lets try the next one :)

Answer 41

this ones a bit easier since the plane is standard and not tilted

Answer 42

okay. point = (3, -1, 1) normal is the yz-plane so that's simply <1, 0, 0> if i'm not mistaken

Answer 43

end up with <3+t, -1, 1>

Answer 44

t = -3

Answer 45

-3 ||1^2|| = -3

Answer 46

yz is the noxplane; how far along the x is our point?

Answer 47

oh, so yz would be <0, 1, 1>?

Answer 48

forget the normal for this; there is really no point with the yz plane to find a normal and work out alot of stuff

Answer 49

think, if a ball is held 5 feet above the floor; how far is it from the floor?

Answer 50

5 feet

Answer 51

right; yz is our floor; and x is our height

Answer 52

whats our height then?

Answer 53

distance between a point and the yz-plane. so the height is just 3

Answer 54

exactly :)

Answer 55

awesome. have time for another, similar, question?

Answer 56

distance between (-1, 3, 2) and 2x-4y+z=1

Answer 57

good, normal anchored to our point

Answer 58

or you can simply enter those parametrics into the plane to get the t

Answer 59

normal = <2, -4, 1>

Answer 60

2x-4y+z double them
t(4-8+2)
t(-2)

2x-4y+z use the point give as
-1 3 2
-------
(-2-12+2) = -12

-12+t(-2) = 1

t = -13/2

Answer 61

<-1 + 2t, 3- 4t,1+t>
2(-1+2t)-4(3-4t)+(1+t) = 1
4t + 16t + t = 14
t = 14/21

Answer 62

okay so i did something wrong here

Answer 63

i feel like that is totally different from what I did on 2a, even though they are similar questions

Answer 64

why is that?

Answer 65

i mean, why did I bother with the parametric form for 2a? is that not required?

Answer 66

-1,3,2 NOT -1,3,1
^^

Answer 67

its required, i just shortcuted the process

Answer 68

ahh

Answer 69

pull the normal and create parametrics that enter back into the equation so the normal just gets ..... ahh, i messed it up to. square them, not double lol

Answer 70

the book says the answer is: \[13/\sqrt{21}\]

Answer 71

i know t = 13 / 21

Answer 72

2x-4y+z SQUARE the normal
t(4+16+1)
t(21)

2x-4y+z use the point give as
-1 3 2
-------
(-2-12+2) = -12

-12+t(21) = 1

t = 13/21

maybe?

Answer 73

where is that sq root on the denominator coming from? it was the same for 2a

Answer 74

its a consequence of form

Answer 75

|n| = ....21 duh, i got that when i squared and added to begin with sqrt(21)

soo

\[\frac{3}{21}\sqrt{21}\]

Answer 76

tell me; is \[\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\]

Answer 77

i don't know. yes?

Answer 78

yes they are equal ( i used my calculator : ) )

Answer 79

\[\frac{1}{\sqrt{2}}*1 =\frac{\sqrt{2}}{2}\]
\[\frac{1}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}} =\frac{\sqrt{2}}{2}\]
\[\frac{\sqrt{2}}{\sqrt{2*2}} =\frac{\sqrt{2}}{2}\]
\[\frac{\sqrt{2}}{2} =\frac{\sqrt{2}}{2}\]

Answer 80

its a result of form, not value ....

Answer 81

ahhhh i see

Answer 82

some books dont want radicals underneath; others dont care

Answer 83

\[13/21 \neq 13/\sqrt{21}\]

Answer 84

correct

Answer 85

\[\frac{13}{\sqrt{21}}=\frac{3\sqrt{21}}{21}\]

Answer 86

i understand the form difference, but the fact remains. so where again is that radical coming from? i feel as though i'm missing a step..

Answer 87

we know t=13/21 right?

and |n| = sqrt(sum of squared parts)

Answer 88

|n| = sqrt(21)

Answer 89

ohhhh i see

Answer 90

awesome

Answer 91

ready for another question?
Compute the area of the parallelogram spanned by the following vectors.
A = <3, -2, 4> B = <5, 1, 1>

Answer 92

sooo, we can shorten this down perhaps bay saying: the distance can be configured as\[something*\frac{|n|}{|n|^2}\]
lol, well at least that will help us eventually

Answer 93

|axb|

Answer 94

find the length of the vector produced from A cross B

Answer 95

in my mind, for distance, it's
1. find t
2. multiply t by the magnitude of the vector

Answer 96

thats a good thing to have :)

Answer 97

oh, and once t is found; we can short it to sqrt(denom)

Answer 98

cool.

for the cross product question: i got -6i + 17j + 13k (or <-6, 17, 13>)