ATAR Notes: Forum

Uni Stuff => Universities - Victoria => University of Melbourne => Topic started by: nubs on March 24, 2012, 08:29:49 pm

Title: Uni Math
Post by: nubs on March 24, 2012, 08:29:49 pm

Thought I may as well make a thread for anyone who needed to ask questions about uni math

I'll start :)
Find the distance of point A from the line L:
A(6,1,2) and L: 2-x=z-2, y=2

I get sqrt(11), but the book says the answer is 3

This is what I did:

I let x=t
L in vector parametric form: (0,2,4) + t(1,0,-1)
I found the vector AL=(-6,1,2) + t(1,0,-1)

So L = ti + 2j + (4-t)k
AL= (t-6)i + j + (2-t)k

Now my reasoning is, the shortest distance between the line and the point is where the point A is perpendicular to the line L, which is where AL is perpendicular to L

So what I then did was the dot product of AL and L, and let this equal 0

I then solved for t, subbed it into AL and found the magnitude for that value of t

What have I done wrong?

Title: Re: Uni Math
Post by: Planck's constant on March 25, 2012, 02:32:51 am

Quote from: Nirbaan on March 24, 2012, 08:29:49 pm

Thought I may as well make a thread for anyone who needed to ask questions about uni math

I'll start :)
Find the distance of point A from the line L:
A(6,1,2) and L: 2-x=z-2, y=2

I get sqrt(11), but the book says the answer is 3

This is what I did:

I let x=t
L in vector parametric form: (0,2,4) + t(1,0,-1)
I found the vector AL=(-6,1,2) + t(1,0,-1)

So L = ti + 2j + (4-t)k
AL= (t-6)i + j + (2-t)k

Now my reasoning is, the shortest distance between the line and the point is where the point A is perpendicular to the line L, which is where AL is perpendicular to L

So what I then did was the dot product of AL and L, and let this equal 0

I then solved for t, subbed it into AL and found the magnitude for that value of t

What have I done wrong?

The book is correct. Again :)

Short solution:

A(6, 1, 2)
L: 2 - x = z - 2, y = 2

* Find the parametric equation of line L

2 - x = t      => x =  2 - t
y = 2
z - 2 = t     => z = 2 + t

Therefore the parametric equation of Line L is,

L: (x, y, z) = (2, 2, 2) + t(-1, 0, 1)

Therefore L passes through the point P(2, 2, 2) and is parallel to the vector v = (-1, 0, 1)

* Form the vector PA

PA = (4, -1, 0)

* All you have to do now is to find the magnitude of the vector resolute of PA = (4, -1, 0) which is perpendicular to vector v = (-1, 0, 1)
As this is bread and butter Spesh 3/4, I'll leave it to you.

The answer is 3

Title: Re: Uni Math
Post by: nubs on March 27, 2012, 07:01:56 pm

ahhhhh yeah you've made me realise that what I did was so very wrong haha

guess that's what happens when you're 4 lectures behind :P

thanks dude!