Given that $\vec{a}$ and $\vec{b}$ are two non-zero vector. The two vectors form 4 resultant vectors such that $\vec{a} + 3\vec{b}$ and $2\vec{a} - 3\vec{b}$ are perpendicular, $\vec{a} - 4\vec{b}$ and $\vec{a} + 2\vec{b}$ are perpendicular. How can I find the angle between $\vec{a}$ and $\vec{b}$?

The answer given here is 114.09. Any help is much appreciated.

  • $\begingroup$ well what do you know about vectors that would give you the angle between them? $\endgroup$ – Sentinel135 Apr 9 '17 at 2:07
  • $\begingroup$ Dot product: a * b = |a| |b| cos @ $\endgroup$ – JF9199 Apr 9 '17 at 2:11
  • $\begingroup$ Alright! Leting $\vec c := \vec a+3\vec b$, $\vec d:= 2\vec a-3\vec b$, $\vec e := \vec a -4\vec b$, and $\vec f := \vec a +2\vec b$, we have $\vec c\cdot \vec d=0$ and $\vec e \cdot \vec f =0$. Now what is the formal definition of the dot product? $\endgroup$ – Sentinel135 Apr 9 '17 at 2:16
  • $\begingroup$ Incidentally, if you write an angle without units, then the units are radians (where $2\pi$ is a full circle). Presumably you mean $114.09^\circ$. $\endgroup$ – Théophile Apr 9 '17 at 3:58

You are given that $\left\langle a+3b,2a-3b\right\rangle=0$ and $\left\langle a-4b,a+2b \right\rangle=0$. Hence

$$\left\langle a+3b,2a-3b\right\rangle=2\left\langle a,a\right\rangle-9\left\langle b,b\right\rangle+3\left\langle a,b\right\rangle=0$$ and $$\left\langle a-4b,a+2b \right\rangle=\left\langle a,a \right\rangle-8\left\langle b,b \right\rangle-2\left\langle a,b \right\rangle=0.$$(Here I assumed that you are working with a real inner-product space). By subtracting the second equation twice from the first, we obtain $$\left\langle b,b \right\rangle+\left\langle a,b \right\rangle=0.$$

Plugging the previous equation into the second yields $$\|a\|^2=6\|b\|^2.$$ Hence $\frac{\|b\|}{\|a\|}=\sqrt{\frac{1}{6}}.$

Now you know that $\left\langle a,b \right\rangle=\cos(\theta)\|a\|\|b\|.$ Thus, after plugging the previous results in the third equation, we get



\begin{cases} (\vec{a}+3\vec{b}).(2\vec{a}-3\vec{b})=0,\\ (\vec{a}-4\vec{b}).(\vec{a}+2\vec{b})=0. \end{cases} $\Rightarrow$ \begin{cases} 2|\vec{a}|^2-9|\vec{b}|^2+3\vec{a}.\vec{b}=0,\\ |\vec{a}|^2-8|\vec{b}|^2-2\vec{a}.\vec{b}=0. \end{cases} $\Rightarrow$ \begin{cases} |\vec{a}|^2=-6\vec{a}.\vec{b},\\ |\vec{b}|^2=-\vec{a}.\vec{b}. \end{cases} then $$\cos\theta=\dfrac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{\vec{a}.\vec{b}}{\sqrt{|\vec{a}|^2|\vec{b}|^2}}=\dfrac{\vec{a}.\vec{b}}{-\sqrt{6}\vec{a}.\vec{b}}=-\dfrac{1}{\sqrt{6}}$$ since $\vec{a}.\vec{b}=-|\vec{b}|^2<0$, so $$\theta=\pi-\arccos\dfrac{1}{\sqrt{6}}=\pi-65^\circ.9=\color{blue}{114^\circ5'41''.4}$$

  • $\begingroup$ At $\cos \theta = 1/6$, how you know $\vec{a}. \vec{b} = 1$ ? $\endgroup$ – JF9199 Apr 9 '17 at 2:47
  • $\begingroup$ It sure looks like $\vec a\cdot\vec b$ is negative, so you've lost a sign. $\endgroup$ – Ted Shifrin Apr 9 '17 at 2:49
  • $\begingroup$ $\vec{a}.\vec{b}=-|\vec{b}|^2<0$. $\endgroup$ – Nosrati Apr 9 '17 at 2:54
  • $\begingroup$ Currently there are two mistakes in the very last two equaltities before the blue answer. $\endgroup$ – Mathematician 42 Apr 9 '17 at 2:57
  • $\begingroup$ Recall that $|x|=\sqrt{x^2}$, not $x=\sqrt{x^2}$ and you forgot the squareroot in the last one. (Also this not critique, just simply pointing out the small errors at the end) $\endgroup$ – Mathematician 42 Apr 9 '17 at 2:59

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