# Dot product questions

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

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

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.

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

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

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

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Galileo
Homework Helper
MiniTank said:
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
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?

Galileo said:
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?
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

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

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

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

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 ?

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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.

dextercioby
Homework Helper
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.

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 $$\sqrt{1^2+(-7)^2+5^2} = 5\sqrt3$$

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

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

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 - $$\theta$$

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tutor69 said:
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 - $$\theta$$
90 - $$\theta$$ since the triangle is right.

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

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

whozum said:
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 $$\sqrt{1^2+(-7)^2+5^2} = 5\sqrt3$$

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

$$cos(\theta) = \frac{a\bullet c}{|A||B|} = \frac{13}{5\sqrt{39/2}} = 0.59$$
how did you get $$cos(\theta) = 0.59$$ i tried it and got 0.83.. and using either 0.83 or 0.59 .. im still getting the wrong answer for $$\Theta$$(34 or 54) i think the answer in the back of the book might be wrong

Whats the answer in the book?

72 degrees and 108 degrees

Its easier to draw than to explain, but this should be easy to understand.
Let it serve as a reminder to all of us that drawing an example always helps.

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thanks for the help whozum. I fell behind in my geometry and discrete class because i missed a week and half because i got my wisdom tooth taken out, so I've been basicly teaching the stuff to myself but I will probably need some more help. Nice to see that people answer so quickly. Thanks again

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