Specialist help

[max payne]

Would appreciate some help with these questions (only the bolded):

5 Let a = 3i − 6j + 4k and b = 2i + j − 2k.
a Find c, the vector component of a perpendicular to b.
b Find d, the vector resolute of c in the direction of a.
c Hence, show that |a| |d| = |c|2.


7) OABC is a parallelogram. A and C are defined by a = 2i + 2j − k and c = 2i − 6j − 3k.
a Find:
i |a − c| ii |a + c| iii (a − c).(a + c)
b Hence, find the magnitude of the acute angle between the diagonals of the
parallelogram.
 
Thanks!

[kamil9876]

Since you have already done 5a and 5b, can't you just plug it in to verify c?

The two diagonals are and . To find the angle between v and w use the dot product:





So now plug in the numbers to find the solution , if it's acute then that's the answer. If not then it is

[Planck's constant]

Since you have already done 5a and 5b, can't you just plug it in to verify c?

The two diagonals are and . To find the angle between v and w use the dot product:





So now plug in the numbers to find the solution , if it's acute then that's the answer. If not then it is

Yes.
The only reason I didnt just think 'whats the problem?' and quickly move on, was the fact that the OP is known to be solid, and I thought there might be some sort of trap.
But there isnt.

Q7 is not a good 'hence' problem. They might as well have done away with part a and gone straight to part b. It would have been the same problem.

Q5 deserves a bit more thought.
At face value it looks like a simple problem of using the vector resolute formulas and crunching the numbers.
And it seems that this was the intend of the problem.
However, this problem could have been made much more interesting if they had not given the values of vectors a and b.
The resulst |a| |d| = |c|2 is always true for all a and b (may be not if they are parallel).
The vectors a, d and c define two similar triangles (easy to see if you draw a diagram of the problem) and the ratio of the corresponding sides is |a|/|c| = |c|/|d|
Therefore, there must be some sort of vector proof to establish the result (left as an exercise for kamil ... hehehe)
 

[max payne]

yeah i realised my problem, I misread the question and drew a ABCD parallelogram, I didnt include O as one of the vertices. As for the other question, just a simple error that didnt allow me to continue. sorry guys Ill try to be more rigid in my checking before posting. thanks

[kamil9876]

nice observation argonaut. Yeah I'm sure you could find a vector proof of the general claim: with , with .

So 

Now since and are parallel

[nina_rox]

Can someone help me with this question?

If f = 2i - j +k and /f-g/ = square root of 6 then g could be..? (/ / means magnitude)

Thank you!

[Planck's constant]

Can someone help me with this question?

If f = 2i - j +k and /f-g/ = square root of 6 then g could be..? (/ / means magnitude)

Thank you!

Interesting.
Since |f| is already sqrt(6), one obvious answer is: g = zero vector

For other solutions, I can geometrically visualise infinite solutions.
Think of f as the line segment AB with length sqrt(6).
Draw a circle of radius sqrt(6) centred B
A lies on the circumference.
Connect A to any other point on the circumference. These are you possible vectors g
Proceeding from here does not look very starightforward to me.

[kamil9876]

The set of all such possible is pretty much the sphere of radius centred at . i.e the set of all such that

[Planck's constant]

The set of all such possible is pretty much the sphere of radius centred at . i.e the set of all such that

perfect

[nina_rox]

Ahh thank you guys!!

[nina_rox]

Wait, for that question would z be (z-1)^2?

And how do I solve it?

Thank you!!

[kamil9876]

Yes, thanks I edited it.

Put then just simplify

[Planck's constant]

Yes, thanks I edited it.

I cant believe you are error prone kamil! I thought you were a machine
As for the OP's question, simply stating what kamil posted IS the solution

[kamil9876]

Nope, if a day goes by with no errors then something is wrong. A possible excuse this time is that I am very rusty with i,j,k notation (hardly ever use it, what if you are working in 27 dimensions? you run out of letters from the alphabet)