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Let $$u = \langle 1, −2, 3 \rangle \qquad \text{and} \qquad v = \langle −4, 5, 6 \rangle.$$ Find the angle between $u$ and $v$, first by using the dot product and then using the cross product.

I used the formula: $U \cdot V = ||u|| \, ||v|| \cos \Delta$ and got $83^\circ$ from the dot product.

However, I am lost as how to use the cross product to find the answer.

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  • $\begingroup$ You will find that the work checks, since $$ \left( \frac{4^2}{14 \cdot 77} \right) \ + \ \left( \frac{1062}{14 \cdot 77} \right) \ = \ 1 \ \ , $$ with $ \ (-27)^2 \ + \ (-18)^2 \ + \ (-3)^2 \ = \ 1062 \ $ as the squared-length of the cross-product vector . $\endgroup$ Jul 30, 2015 at 4:22

2 Answers 2

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Hint The cross product satisfies $$||{\bf a} \times {\bf b}|| = ||{\bf a}|| \, ||{\bf b}|| \sin \theta,$$ where $\theta \in [0, \pi]$ is the angle between $\bf a$ and $\bf b$.

(In fact this property is very nearly one of the common definitions of the cross product; see, e.g., Defn. 7.4 of Dennis G. Zill, Michael R. Cullen (2006). Advanced engineering mathematics (3rd ed.). Jones & Bartlett Learning.)

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    $\begingroup$ ... though there is an ambiguity problem for obtuse angles, which is why we don't usually use cross-products for the purpose of finding the included angle. (That issue does not arise for this particular problem, since the angle is in the first quadrant.) $\endgroup$ Jul 30, 2015 at 4:11
  • $\begingroup$ Yes, once one has the value of $\sin \theta$ in hand, (if it is not equal to $1$) one needs to decide whether the angle is more or less than $\frac{\pi}{2}$, which one can do using, e.g., the dot product. I suppose one could object this isn't "using [only] the cross product", but given the level of the question I suspect that isn't the intention of the question. $\endgroup$ Jul 30, 2015 at 4:16
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    $\begingroup$ Even if it were, one can meet this objection by recalling that we can recover the dot product from the cross product using the formula $${\bf a} \cdot {\bf b} = -\frac{1}{2} \text{tr}({\bf a} \times ({\bf b} \times \, \cdot \,)).$$ $\endgroup$ Jul 30, 2015 at 4:18
  • $\begingroup$ I agree -- this is a typical textbook exercise, but not a method one would generally apply. $\endgroup$ Jul 30, 2015 at 4:19
  • $\begingroup$ Yes, in general there's no reason not to use the dot product, which doesn't suffer the issue you point out, and which anyway works in all dimensions. $\endgroup$ Jul 30, 2015 at 4:20
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Let the angle between $u$ & $v$ be $\alpha$, now the dot product is given as follows $$u\cdot v=|u||v|\cos \alpha$$ $$\implies \cos \alpha=\frac{u\cdot v}{|u||v|}$$ $$=\frac{(1, -2, 3)\cdot (-4, 5, 6)}{|(1, -2, 3)||(-4, 5, 6)|}$$ $$\cos \alpha=\frac{-4-10+18}{\sqrt{(1)^2+(-2)^2+(3)^2}\sqrt{(-4)^2+(5)^2+(6)^2}}$$ $$\cos \alpha=\frac{4}{\sqrt{14\times 77}}$$$$\implies \color{blue}{\alpha=\cos^{-1}\left(\frac{4}{7\sqrt{22}}\right)\approx 83^\circ}$$ Cross product $|u\times v|$ is given as follows $$|u\times v|=\left|\begin{matrix} i&j&k\\ 1&-2&3\\-4&5&6 \end{matrix}\right|$$ $$=|-27i-18j-3k|$$ $$=\sqrt{(-27)^2+(-18)^2+(-3)^2}=\sqrt{1062}=3\sqrt{118}$$ Now, using cross product, the angle $\alpha$ between $u$ & $v$ is given as follows $$u\times v=|u||v|\sin \alpha(\hat n)$$ $$\implies |u\times v|=||u||v||\sin \alpha$$ $$\implies \sin\alpha=\frac{|u\times v|}{|u||v|}$$ $$=\frac{3\sqrt{118}}{7\sqrt{22}}=\frac{3}{7}\sqrt{\frac{59}{11}}$$ $$\color{blue}{\alpha=\sin^{-1}\left(\frac{3}{7}\sqrt{\frac{59}{11}}\right)\approx 83^\circ}$$ Your answer is correct

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  • $\begingroup$ Can you specify i, j, and k in your answer? $\endgroup$
    – shuhalo
    Dec 9, 2016 at 9:58

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