Spec '10 - Help forum

[cipherpol]

Oh, I only calculated the parallel component; the answer also includes the perpendicular component.

They factored out a 2 from the parallel component to get their answer for the first part.

[mandy]

Oh I got it, thank you fady_22 and cipherpol

[mandy]

The points A, B and C have position vectors -i+4j+2k, i-2j-3k, and i-j-2k respectively. Find:
a. the lengths of AB, BC and AC
|AB| =
|BC| =
|AC| =

b. the resolved part of AB in the direction of i-j-2k

I don't know how to do part b of this question, I don't even know what it's asking for. Can someone help me

[cipherpol]

I think they want you to find the component of AB parallel to i-j-2k.

Not sure though :S

[TrueTears]

Yeah cipherpol is spot on.

[mandy]

I think they want you to find the component of AB parallel to i-j-2k.

Not sure though :S

Oh no, I've forgotten what 'the component of AB parallel to i-j-2k' means. What am I meant to do here.

[hyperblade01]

Use the formula

Where a = AB and b = i - j - 2k


EDIT: AB can be broken up in perpendicular and parallel components. Think of a right angled triangle were the hypotenuse is AB and the other two sides are the components - you still end up in the same place you just go a different direction.

[Chavi]

Can I get some help with the following question (specifically the restriction placed on t)

Find the Cartesian equation.
r = (4cost)i + (5sint)j  , t>0

Also, is there a quick way to use mathematical symbols in this forum?
And could somebody explain the concept behind position vectors.

Thanks

[mandy]

Thank you hyperblade01, TrueTears and cipherpol

[TrueTears]

Can I get some help with the following question (specifically the restriction placed on t)

Find the Cartesian equation.
r = (4cost)i + (5sint)j  , t>0

Also, is there a quick way to use mathematical symbols in this forum?
And could somebody explain the concept behind position vectors.

Thanks

Let and




http://vcenotes.com/forum/index.php/topic,10280.0.html


Position vectors are fixed, the line segment must start from the origin. They are NOT free vectors.

[Chavi]

Let and

How about the restriction on t in relation to x

[TrueTears]

domain of cartesian = range of x(t)

range of cartesian = range of y(t)

[Chavi]

domain of cartesian = range of x(t)

range of cartesian = range of y(t)

i.e. |x|<4 ?

[hyperblade01]

[QuantumJG]

The points A, B and C have position vectors -i+4j+2k, i-2j-3k, and i-j-2k respectively. Find:
a. the lengths of AB, BC and AC
|AB| =
|BC| =
|AC| =

b. the resolved part of AB in the direction of i-j-2k

I don't know how to do part b of this question, I don't even know what it's asking for. Can someone help me

let d = i-j-2k 

proj(AB) _d =

I don't like the term 'resolute', what you are doing is projecting one vector onto another (look at this - http://nixweb.com/critters/figs/vproj.gif). Think of w as AB and v as d, this means that vector u is proj(AB) _d .         

Anyway,

proj(AB) _d =  3i-3j-6k