Find vector collinear to sum of other two vectors

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Homework Help Overview

The problem involves finding a vector \(\overrightarrow{c}\) that is collinear with the sum of two vectors \(\overrightarrow{a} + \overrightarrow{b}\), given specific dot products and the magnitude of one of the vectors. The context is within vector algebra and properties of collinearity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to express \(\overrightarrow{c}\) in terms of \(\overrightarrow{a}\) and \(\overrightarrow{b}\) using a scalar multiple. They explore the implications of the given dot products and magnitudes.
  • Some participants suggest using the property of collinearity to relate \(\overrightarrow{c}\) to \(\overrightarrow{a} + \overrightarrow{b}\) and question how to determine the scalar factor.
  • Others propose taking the dot product with \(\overrightarrow{b}\) to derive further relationships and simplify the problem.

Discussion Status

The discussion is ongoing, with participants providing hints and exploring different approaches. The original poster has made attempts to manipulate the equations but expresses difficulty in progressing. Some guidance has been offered regarding the use of dot products to find the scalar multiple.

Contextual Notes

There is a mention of confusion regarding the notation used for the scalar multiple, which may affect clarity in communication. The original poster also notes a perceived error in their initial understanding of the problem setup.

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Homework Statement



Find vector \overrightarrow{c} collinear with the vector \overrightarrow{a}+\overrightarrow{b}, if \overrightarrow{a} \cdot \overrightarrow{b}=5 and \overrightarrow{c} \cdot \overrightarrow{b} = 18, |\overrightarrow{b}|=2

Homework Equations



|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}||\overrightarrow{b}|sin(\overrightarrow{a},\overrightarrow{b})

\mathbf{a}\times\mathbf{b}=\det \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{bmatrix}.

\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}||\overrightarrow{b}|cos(\overrightarrow{a},\overrightarrow{b})

The Attempt at a Solution



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Before I started I found some (I believe) error in the task.

If I subtract the given equations like a.b-c.b=-13 and b(a-c)=-13

|b||a-c|cos(b, a-c) = -13

As we can see on the picture a-c0=-b

|b||a-pc0|cos(b,a-pc0)=-13

Edit: Ok, there isn't error I mixed c0 and c.

Now, let me start.

I approach

c = p(a+b), because c and a+b are collinear, where p is real number.

From the given conditions.

|a||b|cos(a,b)=5

|c||b|cos(c,b)=18

|a|cos(a,b)=5/2

|c|cos(c,b)=18/2

c.(a+b)=|c||a+b|cos(c,a+b)=|c||c0|cos(c,c0)=|c|

Here is where I am stuck...:confused:

II approach

c x (a+b) = 0

c x a + c x b =0

But as you can see I got no coordinates for the vectors, so this is 2nd fail...:mad:

I am dealing with this task for 1 hour and seems like I can't find any way to solve it.

The result in the textbook is c = 2(a+b)

Thanks in advance.

Regards.
 
Last edited:
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Use the simple fact that if two vectors are collinear, one is a scalar multiple of the other. Then you can write

c = k (a + b)

All you need to do is find k. Do you see how to do that in view of the given information?
 
After reading your attempt more carefully, I saw that my hint will not really help you because you've been there. What I call k you call p. So here comes the next hint, what do you get when you dot both sides with b?
 
Last edited:
kuruman said:
After reading your attempt more carefully, I saw that my hint will not really help you because you've there. What I call k you call p. So here comes the next hint, what do you get when you dot both sides with b?

Aaah... I see now.

c.b = (ka+kb).b

18=k(a.b) +k(b.b)

18=5k+4k

9k=18

k=2

Thanks a mill. times, I didn't spot that thing (I thought it was something more complicated).

Regards.
 

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