1. Vectors Application 19

The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus, and determine the lengths of the sides and the angles between the sides.

So I found part A, the lengths of the sides, but I am having trouble finding part B, the angles between the sides.

For part A, the answer I have is [5(3)^0.5]/2, which is correct.

For part B, I think i have to use the formula
cos(theta)=(a.b)/(|a||b|), but when I plug in the given to the formula, I do not get the answer (same as back).

cos(theta)=(6-12+6)/(5(3)^0.5)
theta=90degrees
what did i do wrong?

2. Theorem: A parallelogram is a rhombus if and only if the diagonals are perpendicular.

3. ok.. so how would i find the angle between the lengths of the sides of the rhombus then?

4. Originally Posted by skeske1234
ok.. so how would i find the angle between the lengths of the sides of the rhombus then?
Suppose that $c~\&~d$ are the adjacent vector sides of the rhombus.
We know that $\left\| c \right\| = \left\| d \right\|$. WHY?
$\left\| a \right\| =\sqrt{26}< \left\| b \right\|=7$ therefore $b = c + d\;\& \,a = c - d$. Why is that?

Now you can solve that for $\left\| c \right\|$.

5. so I have
cos(theta)=[a.b]/[7(sqrt(26))]

how do I get a.b without getting 0 as my answer?

what goes on the numerator?

6. or am i using the wrong formula

cos(theta)=(a.b)/(|a||b|)
=(3,-1,-1) . (2,3,-6)
-----------------
7(sqrt(26))
=0
----
7(sqrt(26))

7. Originally Posted by skeske1234
The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus, and determine the lengths of the sides and the angles between the sides.

So I found part A, the lengths of the sides, but I am having trouble finding part B, the angles between the sides.

For part A, the answer I have is [5(3)^0.5]/2, which is correct.

For part B, I think i have to use the formula
cos(theta)=(a.b)/(|a||b|), but when I plug in the given to the formula, I do not get the answer (same as back).

cos(theta)=(6-12+6)/(5(3)^0.5)
theta=90degrees
what did i do wrong?
Did you possibly copy the problem wrong? $(3\vec{i}- 4\vec{j}- \vec{k})\cdot (2\vec{i}+ 3\vec{J}- 6\vec{k})= 3(2)-(4)(3)-(1)(-6)= 6- 12+ 6= 0$ so the two vectors are clearly perpendicular. The angle between the vectors is 90 degrees (or $\pi/2$ radians). What is the answer in your book?

8. The answers they have are 72 degrees and 108 degrees, and no i did not copy the question wrong.. I will re-type without checking above:

The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus and determine the lengths of the sides and the angles between the sides.

9. Originally Posted by skeske1234
The answers they have are 72 degrees and 108 degrees, and no i did not copy the question wrong.. I will re-type without checking above:

The vectors a=3i-4j-k and b=2i+3j-6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus and determine the lengths of the sides and the angles between the sides.
Originally Posted by Plato
Suppose that $c~\&~d$ are the adjacent vector sides of the rhombus.
We know that $\left\| c \right\| = \left\| d \right\|$. WHY?
$\left\| a \right\| =\sqrt{26}< \left\| b \right\|=7$ therefore $b = c + d\;\& \,a = c - d$. Why is that?
Now you can solve that for $\left\| c \right\|$.
From the above you see that $2c=a+b=5i-j-7k$ and $2d=b-a=-i+7j-5k$.

Now that is all need to find $\left\| c \right\|$ and the angle between $c~\&~d$

10. why do you have 2c=... and 2d=... and not c=... and d=...
why is there a 2 in front? how did you get it?

11. Originally Posted by skeske1234
why do you have 2c=... and 2d=... and not c=... and d=...
why is there a 2 in front? how did you get it?
I ask this question, with all due respect, if you have so much trouble with simple algebra then why are you expected to work this problem?

$c = (1/2)\left( {5i - j - 7k} \right) \;\& \,d = (1/2)\left( { - 1i+7j - 5k} \right)$

$\left\| c \right\| = \left\| d \right\| = \sqrt {75/4}$

The angle between them is $\arccos \left( {\frac{{c \cdot d}}{{\left\| c \right\|\left\| d \right\|}}} \right) \approx 72^0$.

Now this is the last time I will answer any question on this thread.

12. In my first answer, I had misread the question- the vectors you are given are the diagonals and they are clearly perpendicular. But you want the angles between the sides. You can get the lengths of the sides as the hypotenuses of right triangles where half the length of the diagonals are the lengths of the legs. Once you know the lengths of the sides, you can find the angles between the sides by using the "cosine law": $c^2= a^2+ b^2- 2ab cos(C)$ where a and b are the lengths of the sides, c is the length of one of the diagonals, and C is the angle opposite that diagonal.