Vector help

[d0minicz]

A, B and C are points defined by position vectors 2i - j -4k, -i + j + 2k and i - 3j - 2k respectively. M is a point of line segment AB such that |AM|=|AC|

 Find:   a)  AM

and

Let a= i -2j + 2k and let b be a vector, such that the vector resolute of a in the direction of b is

a) Find the cosine of the angle between the directions of a and b

b) Find |b| if the vector resolute of b in the direction of a is


thankyou.

[kamil9876]

Haha, i accidentally replied to this question in your other thread about vectors. That's what happens when I have multiple tabs open  :2funny:

http://vcenotes.com/forum/index.php/topic,9556.0.html

[d0minicz]

lol ahaha thanks

need help with the other one now please anyone

[kamil9876]

Refer to diagram attatched.
Vector AC=a  AD=b

The vector resolute a in the direction b is the vector that goes along b and is perpendicular to a. Hence in my diagram vector AB is this vector resolute. But because it is a unit vector, indicated by the hat, it has a magnitude of 1. The length of the hypotenuse AC is actually the magnitude of a. Hence we can easily work out that this length is 3. So now what we have is a right angled triangle, where the hypotenuse AC is 3 units long, and the line AB is 1 unit long.

Hence cos(angle(CAB))=|AB|/|AC|
                               =1/3

[d0minicz]

Thanks !

but is that Cos = |AB|/ |AC| rule come from ?

[kamil9876]

Now for the second one, I'm gonna use the same diagram.

This time: vector AC=b
              vector AD=a

Hence vector AB is the vector resolute of b in the direction a. So vector AB has a magnitude of 2 since it's a unit vector extended by a factor of 2. Now what we want to find is |b|. This is basically the length of the hypotenuse. We can see that:

 cos(angle(CAB))=2/|b|
         |b|=2/cos(angle(CAB))
                       =2/(1/3)
                       =6

[kamil9876]

Thanks !

but is that Cos = |AB|/ |AC| rule come from ?

That's just simple trig. Adjacent over hypotenuse is cos of the angle

[d0minicz]

Oooooooooooopz missed that part  ahahha
thanks alot !

[d0minicz]

 If r = 3i + 3j - 6k,    s= i - 7j +6k  and t = -2i -5j +2k, find the values of and such that the vector r + s + t is parallel to the x-axis.

thanks

[kamil9876]

Vectors that are parallel to the x axis have only an i component e.g: 3i, -1.618i  etc.

I'll call lambda L and I'll call micro M since it's easier to type.

Expand the expression and collect like terms to get:

 r+Ls+Mt=(3+L-2M)i+(3-7L-5M)j+(-6+6L+2M)k

Now the j and k components must be zero and so this yields two equations:

3-7L-5M=0
-6+6L+2M=0

Solve these simultaenously.

[kamil9876]

Oh and just a hint: For the general case, if two vectors a and b are parrellel then a=kb
where k is some real number(other than zero). This general fact could've been used in this situation as well:

Using this we would've got:

i=k(r+Ls+Mt)
          =k((3+L-2M)i+(3-7L-5M)j+(-6+6L+2M)k)
          =k(3+L-2M)i+k(3-7L-5M)j+k(-6+6L+2M)k

Equating the coefficients(note that the left hand side has a coefficient of 0 for the j and k components):

k(3+L-2M)=0
k(-6+6L+2M)=0
Now divide both sides of both equations by k(since k does not equal zero, this can be done) and u get:

3-7L-5M=0
-6+6L+2M=0

Which is what we got earlier

Using this method you can solve a bigger class of problems, namely, if the question said that the vector  r+Ls+Mt was parallel to any non-zero vector. e.g: If it said it was parallel to 2i+3j+4k

[d0minicz]

Hey how do you do these sort of problems:

A unit vector which makes an angle of with the positive direction of the x-axis and an angle of with the positive direction of the y-axis and an obtuse angle with the positive direction of the z-axis is ...?



Another couple ...

The vector ai + 2j + bk is perpendicular to the vector 3i - j + 2k. Find values of a and b?

and

If f = 2i - j + k and |f - g| = then g could be ?

thanks =]





by the way the answers are: 1.
2. a= -2 , b = 4
3.
 

[kamil9876]

First problem: I would say use the dot product and the modulus=1 thing to get three simultaenous equations, and the info that it's obtuse will tell you which solution to take from quadratic(wild guess). Let's see:

Call this vector .





This gives and

using the fact |u|=1 we get:




but since

therefore it's the negative solution

edit: small arithmetic mistake while skipping steps.

[kamil9876]

2.)

3a-2+2b=0

3a+2b=2

There are infinitely many solutions to that equation. (the one provided is just one of them). You sure there is no missing info or is this a multiple choice maybe?

3.) Same story. Infinitely many answers, looks like a multiple choice.

[d0minicz]

yeah question 2 is a multiple choice question; i'll put alternatives:
A. a=1 , b = -1
B. a= 1 , b=0
C. a=-2, b=4
D. a=4 , b=2
E. a=-2, b=-2

3 is also a multiple choice
A. i-2k
B. -i+j
C. -i-j
D. i-k
E. j+k

could u please tell me how to derive ur answer from these alternatives

sorry i didnt put them in the first place =]

thanks ...