What Are Some Dot Product Questions in Geometry and Discrete Class?

[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

 

[TimNguyen]

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

 

[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.

 

[MiniTank]

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

 

[whozum]

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

 

[Data]

Hardly. This board is LaTeX-enabled. Double-click on the images to get a popup with the code to create them. There are lots of samples here:

https://www.physicsforums.com/showthread.php?t=8997

or if you have some time to read, you can use the NSSITL:

http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf

 

[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

 

[Galileo]

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?

 

[MiniTank]

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]

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?

 

[MiniTank]

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

 

[Nylex]

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

 

[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 ?

 

[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.

 

[dextercioby]

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

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

Daniel.

 

[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

 

[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

 

[whozum]

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.

 

[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

 

[MiniTank]

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

U may want to 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.

 

[MiniTank]

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 .. I am still getting the wrong answer for \Theta(34 or 54) i think the answer in the back of the book might be wrong

 

[whozum]

Whats the answer in the book?

 

[MiniTank]

72 degrees and 108 degrees

 

[whozum]

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.

 

[MiniTank]

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