查看完整版本 : Vectors

Vectors

[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture_zpsno6vqb3n.png[/IMG]
[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture1_zpstbt3pumj.png[/IMG]
[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture2_zpsvglpytqq.png[/IMG]
[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture3_zpshmhmyjig.png[/IMG]
[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture5_zpsfp5crgaq.png[/IMG]
[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture4_zpsfuea8vhx.png[/IMG]

Please check if the answer is consistent with the value in the box below each question.

top11 2015-8-30 04:01 AM

[b][u][color=#ff0000]Question 1[/color][/u][/b]

Let the line be
-x = y/3 + 1/3 = z - 2 = t

{x = -t
{y = 3t - 1
{z = t + 2

The line is therefore
(x, y, z) = (-1, 3, 1)t + (0, -1, 2)

Note that (0, -1, 2) is a point on the line.
The direction vector of the line is (-1, 3, 1).

The vector from (0, -1, 2) to (-3, -3, 1) is
(-3, -3, 1) - (0, -1, 2)
= (-3, -2, -1)

The distance between (-3, -3, 1) and (0, -1, 2) is
√[ (-3)² + (-2)² + (-1)²]
= √(9 + 4 + 1)
= √14

[color=#4169e1]Recall that length of the projection of a on b is
| a．b/|b| |[/color]

The length of the projection of the vector (-3, -2, -1) on the line with direction vector (-1, 3, 1) is
| (-3, -2, -1)．(-1, 3, 1)/|(-1, 3, 1)| |
= | (3 - 6 - 1)/√(1 + 9 + 1) |
= 4/√11

Therefore, the distance from (-3, -3, 1) to the line (x, y, z) = (-1, 3, 1)t + (0, -1, 2) is
√[ (√14)² - (4/√11)² ]
= √(14 - 16/11)
= √(138/11)
= [color=#ff0000]3.5419563161414830547172447242969[/color]

回覆2 #的帖子

I got the correct value isn't ?

[url=http://www.discuss.com.hk/android][img=100,23]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img][/url]

Patchouli.K 2015-8-30 08:30 AM

回覆 3# 的帖子

Q1,4,5,6數字正確, Q4的答案還可約簡
Q2,3沒看~ 我不覺得你會錯

top11 2015-8-30 01:02 PM

[quote]原帖由 [i]好男仔[/i] 於 2015-8-30 06:05 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424649197&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
I got the correct value isn't ?
[img]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img] [/quote]
For Question 1, yes, you can see that my final answer is consistent with yours.

For others, I will try to do some later.

回覆 4# 的帖子

Thank you:loveliness:

14c14c 2015-8-30 04:17 PM

[quote]原帖由 [i]top11[/i] 於 2015-8-30 04:01 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424647940&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
Question 1

Let the line be
-x = y/3 + 1/3 = z - 2 = t

{x = -t
{y = 3t - 1
{z = t + 2

The line is therefore
(x, y, z) = (-1, 3, 1)t + (0, -1, 2)

Note that (0, -1, 2) is a point on the l ... [/quote]
one may use differentiation to calculate the minimum
let f(t)= ||(-3,-3,1) - [(-1, 3, 1)t + (0, -1, 2)]||^2 = <(-3,-3,1) - [(-1, 3, 1)t + (0, -1, 2)],(-3,-3,1) - [(-1, 3, 1)t + (0, -1, 2)]>
then one can calculate f'(t) by using product rule d/dt<u(t),v(t)> = <u'(t),v(t)> + <u(t),v'(t)>   (the proof is easy)
and also second derivative indicate that is really a minimum

top11 2015-8-30 04:52 PM

[quote]原帖由 [i]14c14c[/i] 於 2015-8-30 04:17 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424671945&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

one may use differentiation to calculate the minimum
let f(t)= ||(-3,-3,1) - [(-1, 3, 1)t + (0, -1, 2)]||^2 =
then one can calculate f'(t) by using product rule d/dt =    (the proof is easy)
an ... [/quote]
Sure, let's see whether the poster would like to use which method convenient to him/her.
:smile_o12:

回覆7 #的帖子

thanks for suggesting that way as well. but I haven't learned vector calculus yet and all I need to the correct answer!:loveliness:

[url=http://www.discuss.com.hk/android][img=100,23]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img][/url]

回覆8 #的帖子

yup. I didn't expect there's that many people here to help me this time. so thanks:smile_o12:
I will have another sets of questions tomorrow.

[url=http://www.discuss.com.hk/android][img=100,23]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img][/url]

top11 2015-8-30 06:47 PM

[quote]原帖由 [i]好男仔[/i] 於 2015-8-30 06:17 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424678446&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
yup. I didn't expect there's that many people here to help me this time. so thanks:smile_o12:
I will have another sets of questions tomorrow.
[img]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img] [/quote]
What course are you doing?
University Basic Linear Algebra?
of High School A-Level Math?

[quote]原帖由 [i]top11[/i] 於 2015-8-30 06:47 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424680199&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

What course are you doing?
University Basic Linear Algebra?
of High School A-Level Math? [/quote]
For Math, this semester i am doing Calculus 2 and Linear Algebra. I am going to do real analysis next semester and this will probably be my final math subject.:loveliness:

top11 2015-8-30 08:42 PM

[quote]原帖由 [i]好男仔[/i] 於 2015-8-30 08:39 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424687610&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

For Math, this semester i am doing Calculus 2 and Linear Algebra. I am going to do real analysis next semester and this will probably be my final math subject.:loveliness: [/quote]
You mean you are a university student but not major in mathematics?

回覆13 #的帖子

nah. I'm in uni, trying to major in economic and finance

[url=http://www.discuss.com.hk/android][img=100,23]http://i.discuss.com.hk/d/images/r10/androidD.jpg[/img][/url]

Patchouli.K 2015-8-31 07:36 AM

[quote]原帖由 [i]14c14c[/i] 於 2015-8-30 04:17 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424671945&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

one may use differentiation to calculate the minimum
let f(t)= ||(-3,-3,1) - [(-1, 3, 1)t + (0, -1, 2)]||^2 =
then one can calculate f'(t) by using product rule d/dt =    (the proof is easy)
an ... [/quote]

||[(-3,-3,1)-(0, -1, 2)] x (-1, 3, 1)||/||(-1, 3, 1)||

top11 2015-8-31 09:10 AM

[quote]原帖由 [i]Patchouli.K[/i] 於 2015-8-31 07:36 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424718661&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

||[(-3,-3,1)-(0, -1, 2)] x (-1, 3, 1)||/||(-1, 3, 1)||

:smile_o12:

14c14c 2015-8-31 07:36 PM

[quote]原帖由 [i]Patchouli.K[/i] 於 2015-8-31 07:36 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424718661&ptid=25041115][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

||[(-3,-3,1)-(0, -1, 2)] x (-1, 3, 1)||/||(-1, 3, 1)||