Calculate p and q for Multiplying Vectors

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The discussion focuses on calculating the coefficients p and q for the equation \(\overrightarrow{c} = p\overrightarrow{a} + q\overrightarrow{b}\) using three given vectors. Participants emphasize the importance of equating components and solving linear equations derived from the inner products of the vectors. The initial approach is criticized for being mathematically invalid, as it mixes vector and scalar forms incorrectly. The correct method involves using the inner products to create a system of equations that can be solved for p and q. The conversation concludes with the original poster acknowledging the mistake and expressing intent to retry the problem with the provided guidance.
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Three vectors have magnitudes a=3.00m (on the positive x-axis), b=4.00m (makes an angle of 30 degrees with the positive x-axis, counter-clockwise), and c=10.0m (makes an angle of 60 degrees with the negative x-axis, clockwise). If \overrightarrow{c}=p\overrightarrow{a}+q\overrightarrow{b} what are the values of p and q?

I know that the dot product is

\overrightarrow{c}=\overrightarrow{a} \cdot \overrightarrow{b}

So...

\overrightarrow{a} \cdot \overrightarrow{b}=p\overrightarrow{a}+q\overrightarrow{b}

I'm not really sure what my next step should be.
 
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Hint: Draw the 3 vectors, with the head of a touching the tail of b, and the tail of c touching the tail of a. Do you see that you could ratio the sizes of a and b (remember that p & q can be negative numbers if needed), to make the head of the sum of pa+qb meet the head of c?
 
rocophysics said:
\overrightarrow{c}=\overrightarrow{a} \cdot \overrightarrow{b}\overrightarrow{a} \cdot \overrightarrow{b}=p\overrightarrow{a}+q\overrightarrow{b}

Those are not even mathematically valid equations! In the first one, you have a vector on the left and a scalar on the right. (Remember, the dot product of two vector results in a scalar.) In the second, it's the other way around.

The simplest way, if you ask me, is to equate components and solve the resulting set of linear equations.
 
rocophysics said:
\overrightarrow{c}=p\overrightarrow{a}+q\overrightarrow{b}
I'm not really sure what my next step should be.
As neutrino noted, your approach is not valid. You are on the right track, however. Taking the inner product with \vec a and \vec b yields a pair of linear equations in p and q:

<br /> \begin{aligned}<br /> \vec a \cdot \vec c &amp;= \vec a \cdot \vec a \, p + \vec a \cdot \vec b \, q \\<br /> \vec b \cdot \vec c &amp;= \vec b \cdot \vec a \, p + \vec b \cdot \vec b \, q<br /> \end{aligned}<br />

You have the requisite information needed to determine all of the inner products in the above equations. Solve the system of equations for p and q, and voila, you have the answer.
 
neutrino said:
Those are not even mathematically valid equations!

D H said:
<br /> \begin{aligned}<br /> \vec a \cdot \vec c &amp;= \vec a \cdot \vec a \, p + \vec a \cdot \vec b \, q \\<br /> \vec b \cdot \vec c &amp;= \vec b \cdot \vec a \, p + \vec b \cdot \vec b \, q<br /> \end{aligned}<br />
Can't believe I made such a dumb mistake. Alright, I will attempt this problem again with your suggestions, thanks!
 
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