Need help with this Vector Question +rep

[simpleas123]

Alright all I am on this vector question and I am having problems with this question. Part (i) i have no idea what to start with / how i possibly prove the question.

This is the question, help will be deeply appriciated. Thanks, David. Also note that
|| a || = (a_1 + a_2 + ... + 1_n)^1/2 .

|| a - b || = || a || || b || - 2 (a \cdot b).

I have done part (i) :) Its the rest I am struggling with.

 

[berkeman]

Could you show us how you solved part (i)?

 

[simpleas123]

Could you show us how you solved part (i)?

c = a + t(b-a), take t = l1 / l
c = a + (l1 / l)(b-a), Expand
c = a -a(l1 / l) + b(l1 / l) Collect terms with a.
c = a(1 - (l1 / l)) + b(l1 / l) 1 = l / l
c = a(l/l - (l1 / l)) + b(l1 / l)
c = a(l - l1)/l + b(l1 / l) l - l1 = l2
c = a(l2 / l) + b(l1 / l)

Its the rest I am stuck on :(
Any ideas?

 

[gabbagabbahey]

Hi simpleas123, welcome to PF!

c = a + t(b-a), take t = l1 / l

You should probably use boldface type to distinguish vectors from scalars, just to make things clear (Clicking on the "go advanced" button will give you all kinds of typesetting options, including using $\LaTeX$)

Also, although $t=\frac{l_1}{l}$ is correct, you need to provide a justification of this for your proof to be complete (A justification of the hint given is also necessary).

Its the rest I am stuck on :(
Any ideas?

For problem (ii), proving the statement "If $\textbf{x}=\textbf{y}$, then $\textbf{c}\cdot\textbf{x}=\textbf{c}\cdot\textbf{y}$ and $\textbf{c}\times\textbf{x}=\textbf{c}\times\textbf{y}$" should be easy enough...as for proving the opposite direction if-then statement; what is the result of problem 3 (b) they mention?...What is property (4) of the cross product?

 

[simpleas123]

Property 4 is the same as result of problem 3 (b).
It states that;

Where a,b are vectors

a x b = 0, if and only if there exist two scalars $$\mu,\lambda$$ one of which is non-zero such that a$$\lambda$$ = b$$\mu$$

I have done part ii) now too.

 

[gabbagabbahey]

Okay, and what was your solution for (ii)?

Problem (iii) appears to contain a typo...I am fairly certain it's suppose to say:

"Show that the equation of the line $l$ through the points $\textbf{x}_0$ and $\textbf{x}_0+\textbf{a}$ is $(\textbf{x}-\textbf{x}_0)\times\textbf{a}=0$"

To do this, just use what property 4 tells you about the equation $(\textbf{x}-\textbf{x}_0)\times\textbf{a}=0$...

 

[simpleas123]

Okay, and what was your solution for (ii)?

Problem (iii) appears to contain a typo...I am fairly certain it's suppose to say:

"Show that the equation of the line $l$ through the points $\textbf{x}_0$ and $\textbf{x}_0+\textbf{a}$ is $(\textbf{x}-\textbf{x}_0)\times\textbf{a}=0$"

To do this, just use what property 4 tells you about the equation $(\textbf{x}-\textbf{x}_0)\times\textbf{a}=0$...

Part ii)
$c \cdot y = c \cdot x$
$(c \cdot y) - (c \cdot x) = 0$
$c \cdot (x-y) = 0$

$\therefore y = x$
and it was pretty much the same method for the cross product execpt u had to state a property at the end as x x y could equal something other than 0.

 

[gabbagabbahey]

Part ii)
$c \cdot y = c \cdot x$
$(c \cdot y) - (c \cdot x) = 0$
$c \cdot (x-y) = 0$

$\therefore y = x$
and it was pretty much the same method for the cross product execpt u had to state a property at the end as x x y could equal something other than 0.

Careful, $\textbf{c}\cdot(\textbf{x}-\textbf{y})=0$ doesn't necessarily mean that $\textbf{x}-\textbf{y}=0$...one of the vectors in a dot product being zero is not the only way for the dot product to be zero...

 

[Mark44]

$\bold{c} \cdot (\bold{x}-\bold{y}) = \bold{0}$
This doesn't necessarily imply that x = y. All it says is that c is orthogonal to x - y.

Counterexample: c = (1, 0, 0), x = (0, 2, 0), y = (0, 1, 0).
$\bold{c} \cdot (\bold{x}-\bold{y}) = (1, 0, 0) \cdot (0, 1, 0) = \bold{0}$
but x - y isn't 0.

 

[simpleas123]

$\bold{c} \cdot (\bold{x}-\bold{y}) = \bold{0}$
This doesn't necessarily imply that x = y. All it says is that c is orthogonal to x - y.

Counterexample: c = (1, 0, 0), x = (0, 2, 0), y = (0, 1, 0).
$\bold{c} \cdot (\bold{x}-\bold{y}) = (1, 0, 0) \cdot (0, 1, 0) = \bold{0}$
but x - y isn't 0.

Oh snap that's right, well then how would i prove it?

 

[gabbagabbahey]

Oh snap that's right, well then how would i prove it?

Well, $\textbf{c}\cdot(\textbf{x}-\textbf{y})=0$ tells you either $\textbf{x}-\textbf{y}=0$ or $\textbf{x}-\textbf{y}$ is orthogonal to $\textbf{c}$...right?

What can you say about $\textbf{c}\times(\textbf{x}-\textbf{y})$ if $\textbf{x}-\textbf{y}$ is orthogonal to $\textbf{c}$?