Dot product questions

  • Thread starter MiniTank
  • Start date
  • #1
62
0
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
 

Answers and Replies

  • #2
80
0
Okay... did you make any attempt at the problems?
 
  • #3
2,209
1
1a)[tex] X dot Y = |X||Y|cos(\theta) [/tex]


#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 = [tex]X_iY_i + X_jY_j + X_kY_k [/tex]

and X dot Y = [tex]|X||Y|cos(\theta)[/tex]

Where Y is your third vector, and X is one of your original vectors.
 
  • #4
62
0
how do you create those formulas .. do you copy them from word?
 
  • #5
2,209
1
Use the forum's latex capabilities. The tags are [ tex ] [/tex]
 
  • #7
62
0
heres what i tried

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


not sure what to do next

#2) did the dot product... gave me zero
did a-c .. got (1,-7,5) .. did [tex] \sqrt{1^2+-7^2+5^2} = \sqrt[5]{3} [/tex]
the back of the book .. has [tex] \frac{\sqrt[5]{3}}{2} [/tex] ... why is it over 2?
then for one of the angles i did this:
[tex] cos\Theta = \frac{\vec{a}\bullet\vec{c}}{|\vec{a}||\vec{c}|} [/tex]
[tex] cos\Theta = \frac{26}{\sqrt{1950}} [/tex]
[tex] \Theta = 54 [/tex]
which is wrong
 
Last edited:
  • #8
Galileo
Science Advisor
Homework Helper
1,989
6
MiniTank said:
heres what i tried

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


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?
 
  • #9
62
0
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
 
  • #10
Galileo
Science Advisor
Homework Helper
1,989
6
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 [itex]\vec a[/itex] and [itex]\vec b[/itex]?
 
  • #11
62
0
theyre 1? .. then youll end up getting 11/2 = 4 .. which ends up giving you 3/2 right?
 
  • #12
551
1
Yeah, |a| = |b| = 1 cos you're told they're unit vectors. 11/2 isn't 4 though :confused:.
 
  • #13
15
0
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:
  • #14
2,209
1
6i+j dot i-2j theta = 60

mag a = [tex] \sqrt{6^2+1^2} = sqrt{37} [/tex]

mag b = [tex] \sqrt{1^2+2^2} = sqrt{5} [/tex]

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

Sorry for the different variables.
 
  • #15
dextercioby
Science Advisor
Homework Helper
Insights Author
13,005
554
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
2,209
1
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

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

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

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

[tex] cos(\theta) = \frac{a\bullet c}{|A||B|} = \frac{13}{5\sqrt{39/2}} = 0.59 [/tex]
 
  • #17
15
0
The answer in book is right. The length of side1, [tex]|\vec{X}|[/tex] is obtained by

[tex] \vec{X} = \frac {\vec{a} - \vec{b}}{2} [/tex]

The length of side 2 is, [tex]|\vec{Y}|[/tex]

[tex] \vec{Y} = \frac {\vec{a} + \vec{b}}{2} [/tex]

The angle, [tex] \theta [/tex] between two sides can be found by X.Y and the other angle would be = 180 - [tex] \theta [/tex]
 
Last edited:
  • #18
2,209
1
tutor69 said:
The answer in book is right. The length of side1, [tex]|\vec{X}|[/tex] is obtained by

[tex] \vec{X} = \frac {\vec{a} - \vec{b}}{2} [/tex]

The length of side 2 is, [tex]|\vec{Y}|[/tex]

[tex] \vec{Y} = \frac {\vec{a} + \vec{b}}{2} [/tex]

The angle, [tex] \theta [/tex] between two sides can be found by X.Y and the other angle would be = 180 - [tex] \theta [/tex]
90 - [tex] \theta [/tex] since the triangle is right.
 
  • #19
15
0
We are finding the angles between the two sides or say the angles that makes the two corners so it should be 180 - [tex] \theta[/tex]
 
  • #20
62
0
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.
 
  • #21
62
0
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

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

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

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

[tex] cos(\theta) = \frac{a\bullet c}{|A||B|} = \frac{13}{5\sqrt{39/2}} = 0.59 [/tex]
how did you get [tex] cos(\theta) = 0.59[/tex] i tried it and got 0.83.. and using either 0.83 or 0.59 .. im still getting the wrong answer for [tex]\Theta[/tex](34 or 54):frown: i think the answer in the back of the book might be wrong
 
  • #22
2,209
1
Whats the answer in the book?
 
  • #23
62
0
72 degrees and 108 degrees
 
  • #24
2,209
1
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.
 

Attachments

Last edited:
  • #25
62
0
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 :approve:
 
Last edited:

Related Threads on Dot product questions

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
524
  • Last Post
Replies
2
Views
915
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
4
Views
822
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
10
Views
2K
Top