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The exercise is such: Given that $|\vec{a}| = 3$, $|\vec{b}| = 2$ and $\varphi = 60^{\circ}$ (the angle between vectors $\vec{a}$ and $\vec{b}$), calcluate scalar product $(\vec{a}+2\vec{b}) \cdot (2\vec{a} - \vec{b})$.

My initial thought was to solve the requested product by "sticking" separately calculated fragments together and then follow the definition of scalar product, which is

$$\vec{a} \cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos\varphi.$$

Via observation

$$\vec{a} \cdot \vec{a} = |\vec{a}| \cdot |\vec{a}| \cdot \cos0^{\circ} = |\vec{a}|^2 \Longrightarrow |\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$$

we can get values of $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$ (mind the notation: $a^2$ in this case represents $a^2 = |\vec{a}|^2$, and $b^2 = |\vec{b}|^2$):

\begin{align*} |\vec{a}+2\vec{b}| &= \sqrt{(\vec{a}+2\vec{b}) \cdot (\vec{a}+2\vec{b})} =\sqrt{a^2 + 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\ &= \sqrt{3^2 + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2}+2^2} = 5. \end{align*}

Similarly,

\begin{align*} |2\vec{a} - \vec{b}| &= \sqrt{(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b})} = \sqrt{4a^2 - 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\ &=\sqrt{4 \cdot 3^2 - 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} + 2^2} = 2\sqrt{7}. \end{align*}

Now we plug both results in:

$$(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) = |\vec{a} + 2\vec{b}| \cdot |2\vec{a} - \vec{b}| \cdot \cos60^{\circ} = 5 \cdot 2\sqrt{7} \cdot \frac{1}{2} = 5\sqrt{7}.$$

However, this is not the right solution according to my textbook. The correct result is $19$. I thought about it for a bit and took a different route:

\begin{align*} (\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) &= \vec{a} \cdot 2\vec{a} - \vec{a} \cdot \vec{b} + 4 \vec{a} \cdot \vec{b} - 2\vec{b} \cdot \vec{b} = \\ &=2a^2 - |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + 4 |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} - 2b^2 = \\ &= 2 \cdot 3^2 - 3 \cdot 2 \cdot \frac{1}{2} + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} - 2 \cdot 2^2 = 19. \end{align*}

The second method clearly worked, while the first one failed miserably. But my question is why did my first approach fail? Did I get the wrong results when calculating $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$? I have no idea. Please, help me understand my mistakes. Thank you in advance.

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3 Answers 3

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In the first method you assume at the end that the angle between $\vec{(a+2b)}$ and $\vec{(2a-b)}$ is $60^\circ$. If you were going to do this approach you would need the angle between them instead of $\cos 60^\circ$ for the last line.

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I think it is failed because here

$(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) = |\vec{a} + 2\vec{b}| \cdot |2\vec{a} - \vec{b}| \cdot \cos60^{\circ} = 5 \cdot 2\sqrt{7} \cdot \frac{1}{2} = 5\sqrt{7}$

you are assuming that the angle between vectors $(\vec{a} + 2\vec{b})$ and $(2\vec{a} - \vec{b})$ is $60^\circ$ but it may not be (and apparently it is not).

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Note $$2\vec {b}\cdot 2\vec {b}=4b^2$$ Compare this with what you have in your first approach when calculating $$ |\vec {a}+2\vec {b}| $$

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  • $\begingroup$ Splendid. I now have the right solution. Thank you and the other two that pointed out my mistake in intuition. $\endgroup$
    – God bless
    Commented Dec 24, 2017 at 15:11
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    $\begingroup$ This is one error in the calculation, but I think the error pointed out by other answers is more fundamental, because it shows that the entire idea of what you are trying to do here is wrong. Try putting $4b^2$ instead of $b^2$ in the calculation of $|\vec{a}+2\vec{b}|$ and see if you come up with the right answer to the problem. $\endgroup$
    – David K
    Commented Dec 24, 2017 at 15:53
  • $\begingroup$ @DavidK yes that is indeed important $\endgroup$
    – PM.
    Commented Dec 24, 2017 at 16:16
  • $\begingroup$ @DavidK I already thanked and upvoted the other two guys for pointing out my mistake in intuition; since two people did it, I had no choice but to give a tick to the answer that pointed out a new unique error. $\endgroup$
    – God bless
    Commented Dec 24, 2017 at 16:40
  • $\begingroup$ You did acknowledge the other two answers in your comment (as I just noticed--my fault for reading it too quickly the first time). I suppose the second half of my comment was unnecessary, sorry! $\endgroup$
    – David K
    Commented Dec 24, 2017 at 16:46

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