# Dot product questions

1. Mar 28, 2005

### MiniTank

I have some questions from my geometry and discrete class..

#1 Given a and b unit vectors,
a) if the angle between them in 60 degrees, calculate (6a+b)●(a-2b)
b) if |a+b|= root(3), determine (2a-5b)●(b+3a)

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

** for number 2) a and b are vectors ... i,j,k are unit vectors

2. Mar 28, 2005

### TimNguyen

Okay... did you make any attempt at the problems?

3. Mar 28, 2005

### whozum

1a)$$X dot Y = |X||Y|cos(\theta)$$

#2. Dot the two vectors together to make sure they are perpendicular. You should get 0. Subtracting the two vectors should give you a third vector representing the side.

Then to find the angles, use the relationship:

X dot Y = $$X_iY_i + X_jY_j + X_kY_k$$

and X dot Y = $$|X||Y|cos(\theta)$$

Where Y is your third vector, and X is one of your original vectors.

4. Mar 28, 2005

### MiniTank

how do you create those formulas .. do you copy them from word?

5. Mar 28, 2005

### whozum

Use the forum's latex capabilities. The tags are [ tex ] [/tex]

6. Mar 28, 2005

### Data

7. Mar 28, 2005

### MiniTank

heres what i tried

1a) $$(6\hat{a}+\hat{b})\bullet(\hat{a}-2\hat{b})$$
$$= 6|\hat{a}|^2 - 12\hat{a}\bullet\hat{b} + \hat{b}\bullet\hat{a} - 2|\hat{b}|^2$$
$$= 6|\hat{a}|^2 - 11\hat{a}\bullet\hat{b}- 2|\hat{b}|^2$$
$$= 6|\hat{a}|^2 - 11|\hat{a}| |\hat{b}|cos\Theta- 2|\hat{b}|^2$$
$$= 6|\hat{a}|^2 - 11|\hat{a}| |\hat{b}|cos60- 2|\hat{b}|^2$$
$$= 6|\hat{a}|^2 - 11|\hat{a}| |\hat{b}|(1/2)- 2|\hat{b}|^2$$

not sure what to do next

#2) did the dot product... gave me zero
did a-c .. got (1,-7,5) .. did $$\sqrt{1^2+-7^2+5^2} = \sqrt[5]{3}$$
the back of the book .. has $$\frac{\sqrt[5]{3}}{2}$$ ... why is it over 2?
then for one of the angles i did this:
$$cos\Theta = \frac{\vec{a}\bullet\vec{c}}{|\vec{a}||\vec{c}|}$$
$$cos\Theta = \frac{26}{\sqrt{1950}}$$
$$\Theta = 54$$
which is wrong

Last edited: Mar 28, 2005
8. Mar 28, 2005

### Galileo

Okay, you'd know the answer if you knew what |a| and |b| are right?
So can you find out what they are from the given data?

9. Mar 28, 2005

### MiniTank

im assuming your talking about |a + b| = root 3 ... which is only used for b) .. not a) .. so you DONT know what |a| and |b| are

10. Mar 28, 2005

### Galileo

No, I'm talking about question a). What can you say about the magnitudes of $\vec a$ and $\vec b$?

11. Mar 28, 2005

### MiniTank

theyre 1? .. then youll end up getting 11/2 = 4 .. which ends up giving you 3/2 right?

12. Mar 28, 2005

### Nylex

Yeah, |a| = |b| = 1 cos you're told they're unit vectors. 11/2 isn't 4 though .

13. Mar 28, 2005

### tutor69

Now since it is given that a and b are unit vectors , therefore |a| = 1 and |b|=1

so (6a + b).(a-2b) = 4 - (11/2) = -3/2

Now are you able to get the result ?

Last edited: Mar 28, 2005
14. Mar 28, 2005

### whozum

6i+j dot i-2j theta = 60

mag a = $$\sqrt{6^2+1^2} = sqrt{37}$$

mag b = $$\sqrt{1^2+2^2} = sqrt{5}$$

a dot b = $$|a||b|cos(\theta) = sqrt{185}(1/2)$$

Sorry for the different variables.

15. Mar 28, 2005

### dextercioby

If u're using Latex to edit your formulas,then use the function "\frac{...}{...}" for writing fractions...

U may wanna pay more attention with paranthesizing when necessary (v.post #7,the square roo t with 2 consecutive signs & no round brackets).

Daniel.

16. Mar 28, 2005

### whozum

For #2, remember that vectors A and B and the adjoining vector C represent one quarter of the rhombus. A diagonal stretches across half of a rhombus, before you subtract the two vectors and dot them, you need to half them.

a dot c

$$\hat a = 3/2\hat i - 2\hat j - 1/2\hat k$$ with Magnitude: $$\sqrt{3/2^2+(-2)^2+(-1/2)^2} = \sqrt{9/4+4+1/4} = \sqrt{13/2}$$

$$\hat c = 1/2\hat i - 7/2\hat j + 5/2\hat k$$ with Magnitude $$\sqrt{1^2+(-7)^2+5^2} = 5\sqrt3$$

$$a \bullet c = ((3)(1) + (-4)(-7) + (-1)(5))(1/2) = (1/2)(3 + 28 -5) = 13$$

$$cos(\theta) = \frac{a\bullet c}{|A||B|} = \frac{13}{5\sqrt{39/2}} = 0.59$$

17. Mar 28, 2005

### tutor69

The answer in book is right. The length of side1, $$|\vec{X}|$$ is obtained by

$$\vec{X} = \frac {\vec{a} - \vec{b}}{2}$$

The length of side 2 is, $$|\vec{Y}|$$

$$\vec{Y} = \frac {\vec{a} + \vec{b}}{2}$$

The angle, $$\theta$$ between two sides can be found by X.Y and the other angle would be = 180 - $$\theta$$

Last edited: Mar 28, 2005
18. Mar 28, 2005

### whozum

90 - $$\theta$$ since the triangle is right.

19. Mar 28, 2005

### tutor69

We are finding the angles between the two sides or say the angles that makes the two corners so it should be 180 - $$\theta$$

20. Mar 28, 2005

### MiniTank

i just put the wrong type of slash before the sqrt, i'll edit it, but i think everyone understood what i did.