1. ## Rhombus (vectors)

If x= 3i -4j-k and y=[3,2,-1], are diagonals of a parrallelagram, then show that this parrarellagram is a rhombus.

So, I'm assuming I would have to prove that the magnitudes of both y and x are equal. But I'm a bit confused with this unit vector deal. (i,j,k)

2. First, the "unit vector deal" you're having problems with is just notation; in your example, x is he same as [3, -4, -1].

You're not quite right about hoping they have equal magnitutes. A parallelogram with equal-lengthed diagonals is a rectangle, not a rhombus. A rhombus has perpendicular diagonals, so they want you to take the dot product and show that it equals zero.

Speaking of which, I'm guessing you copied the problem down wrong: either x=3i-4j+k, or y=[3,2,+1].

Hope this helps.

3. "Speaking of which, I'm guessing you copied the problem down wrong"

Yes you are right. The question should read that y=[2,3,-6]

4. That'll work out just fine.

5. Ok, I was confusing the sides with the diagonals. Thanks for your help.

6. Now, how could I go about trying to find the length of the sides of the rhombus?

7. There is something wrong with the way you have given the problem.

It is well known that: A parallelogram is a rhombus if and only if its diagonals are perpendicular.
So see if their dot product is zero.

8. "There is something wrong with the way you have given the problem.

It is well known that: A parallelogram is a rhombus if and only if its diagonals are perpendicular.
So see if their dot product is zero."

The question asked to prove that it was a rhombus, which Henderson realized could not be done as I gave the wrong numbers to begin with.

9. Originally Posted by Mtl
The question asked to prove that it was a rhombus, which Henderson realized could not be done as I gave the wrong numbers to begin with.
That is the point.
If the dot product $\displaystyle x \cdot y \ne 0$ then it is not a rhombus.
Either the problem is wrong or you have copied it incorrectly.

10. Ok, I'm still confused here... Wouldn't the diagonals' magnitude have to be equal in a rhombus?

11. Originally Posted by Mtl
Ok, I'm still confused here... Wouldn't the diagonals' magnitude have to be equal in a rhombus?
No, that is true of rectangles. It is not necessary in a rhombus.
A rhombus with congruent diagonals is a square.

12. Oh ok... So then to determine the length of the sides of the rhombus I would just divide the magnitudes of the two diagonals by 2 and then use pythagorum's theorum?

13. Originally Posted by Mtl
If x= 3i -4j-k and y=[3,2,1], are diagonals of a parallelogram, then show that this parallelogram is a rhombus.
I have a hard time understanding why you need the length of the sides.
Note I have changed one sign in the problem. Now $\displaystyle x \cdot y = \left\langle {3, - 4, - 1} \right\rangle \cdot \left\langle {3,2,1} \right\rangle = 0$.
Now that alone tells us the parallelogram with diagonals x & y is a rhombus.
There is no need to find the lengths of the sides.
The problem does not call for it.

14. Sorry for the confusion, but the original question was indeed copied down wrong by me it should have read:
If x= 3i - 4j -k and y= [2,3,-6], are the diagonals of a parrallelgram then,
(a) show that the parralelagram is a rhombus
(b) find the length of the sides of the rhombus

*I have already proved part A through a dot product but I'm not sure as to how to figure out part B.

15. Well, we are not mind readers.
If s is a side of the rhombus then
$\displaystyle \left\| s \right\|^2 = \left\| {.5x} \right\|^2 + \left\| {.5y} \right\|^2$

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