vectors!!!! :(

[yang_dong]

question 9a) plz....

isn't b = pi - qj?
c= -pi +qj?

thanks you

[keltingmeith]

Note that all we're doing here is ROTATING the vector - so their magnitudes stay the same. There are a couple of ways to handle this question, but let's consider some things:

i) These vectors are two dimensional, and so are very similar to complex numbers
ii) Each rotation is by 90 degrees (pi/2) - the same as multiplying a complex number by i

So, how about we take the vector and write it in complex form .

This means that the points given by b and c on the complex plane are done by multiplying by i once (90 degrees counter-clockwise) and three times (270 degrees counter-clockwise/90 degrees clockwise).

This gives us:




Now, all you need to do is simplify these expressions and convert them back into vector notation.

You also could've done this transformation with matrices, but that's first year linear algebra.

[yang_dong]

oh ok thank you...

its just cause we haven't been taught complex yet.

and um also.. and ans are
b = qi - pj
c = -qi +pj,
the q and p have swapped!!

[keltingmeith]

That is troublesome... Welp, something you COULD do is this:

Know that if you turn the vector into a triangle, you can see that the points can be found in terms of , where:




and is the angle made by the vector and the x-axis.

So, if we rotate 90 degrees clockwise, that's the same as taking away 90 degrees from the angle:



Using symmetry properties, you'll get:



Doing likewise for point c (except adding 90 degrees instead) gives:



And then, knowing that and , you should be able to find your new vectors.

However, I reckon once you've learned complex numbers, the first method I gave you would be better to use.

[yang_dong]

thank you...

if i use the complex method you taught me... how do i know which is the i component and which is the j component?

[keltingmeith]

This is something you'll hopefully pick up when you get to complex numbers. When you learn about Argand planes, you'll notice that Re(z) is the same as the x-axis on a cartesian plane, and Im(z) is the same as the y-axis on a cartesian plane. Basically, this means that all real parts will turn into i-components, and all imaginary parts will turn into j-components.

Note that to use my method, you'll also know how to simplify , which admittedly isn't too hard, but you'll need to know what you're doing.

[yang_dong]

There is this question: Points A, B, C and D defined by position vectors a, b,c and d respectively. AB +CD = 0
how would i prove that ABCD is a rhombus if magnitude a = magnitude c and angles AOB and BOC are equal?

thanks you

[asdfqwerty]

another question here as well please
A i-j+k
B 11i-j+k
D i+5j+9k
ABCD is a square

Find the co ordinates of C