# Angle between a vector and cross product of two vectors

The angle between the vectors $$\overrightarrow {a}$$ and $$\overrightarrow {b}$$ is $$\pi/3$$, $$\overrightarrow {b}$$ and $$\overrightarrow {c}$$ is $$\pi/4$$, $$\overrightarrow {c}$$ and $$\overrightarrow {a}$$ is $$\pi/6$$. Find the angle between $$\overrightarrow {a}$$ and $$\overrightarrow {b}\times \overrightarrow {c}$$

I tried to calculate the angle by drawing the diagram and geometry but it didn't work out for me. The brute Force method of fixing two of the vectors as convenient position vectors also got messy.

How can we evaluate this? I would prefer a general approach, but if there isn't any, a brute Force approach would also be fine.

We only need angles so assume all vectors are unit vectors. Angles between vectors are ab , bc and ca .

Now $$\sin\theta=\frac{\left|\overrightarrow a \times(\overrightarrow b \times \overrightarrow c)\right|}{\left|\overrightarrow a\right|\cdot|b\times c|}=\frac{\left|(a\cdot c)\overrightarrow b-(a\cdot b)\overrightarrow c\right|}{|a|\cdot|b|\cdot|c|\cdot\sin bc}$$ $$=\frac{\left|\overrightarrow b \cos ac-\overrightarrow c \cos ab\right|}{\sin {\pi \over 4}}$$ $$=\frac{\sqrt{1^2\cdot\cos^2 ac+1^2\cdot \cos^2 ab-2(\cos ac)(\cos ab)(\cos ac)}}{{1 \over \sqrt2}}$$

Can you take it from here.

Assume WLOG unitary vectors and $$\vec a=\hat i$$ and $$\vec b\in (x,y)$$ plane, then we have that

• $$\hat i\cdot \vec b =\frac12 \implies b_x=\frac12,\; b_y=\frac{\sqrt 3}2$$
• $$\vec b\cdot \vec c =\frac{\sqrt 2}2\implies \frac12 c_x+\frac{\sqrt 3}2c_y=\frac{\sqrt 2}2 \implies c_x+\sqrt 3 c_y=\sqrt 2$$
• $$\vec c\cdot \hat i =\frac{\sqrt 3}2 \implies c_x=\frac{\sqrt 3}2 \implies c_y=\frac{\sqrt 2}{\sqrt 3}-\frac{1}{2}$$

then we can find $$c_z$$ and finally the requested angle.

Here is a plot for this particular solution

• $\vec b\cdot \vec c =\frac{\sqrt 2}2$ is only true if $\;|b|·|c|=1 \;$ which is not in the given data. Same goes for the others modules. Sep 27, 2019 at 16:24
• @Ripi2 I'm assuming wlog unitary vectors.
– user
Sep 27, 2019 at 16:25
• Nice to see you are back and active again. $c_z=\sqrt{\sqrt{2/3}-2/3}$. Then would not $b\times c$ and $|b\times c|$ be tedious? Any shortcuts? Sep 27, 2019 at 16:51
• @farruhota Thanks, nice to see you again! I've just added some elaboration by wolfy. I obtain a quite different $c_z$.
– user
Sep 27, 2019 at 16:53
• $c_a^2+c_b^2+c_z^2=1 \Rightarrow c_z=\sqrt{1-(\sqrt3/2)^2-(\sqrt{2/3}-1/2)^2}= \sqrt{\sqrt{2/3}-2/3}$. You are right, it is equal to yours. Sep 27, 2019 at 16:55

Here is one solution, which is quite general:

Assuming the vectors are 3D with right handed orthonormal basis $$(\hat i,\hat j, \hat k)$$ :

Without loss of generality, fix $$\overrightarrow a$$ as $$a \hat i$$ and $$\overrightarrow b$$ as $$b(\cos\frac\pi 3 \hat i + \sin\frac\pi 3 \hat j) = b(\frac 12 \hat i + \frac{\sqrt{3}} 2 \hat j)$$. We know $$c$$ must be of the form $$c(cos\frac\pi6 \hat i + s \hat j + t \hat k)=c(\frac {\sqrt3}2 \hat i + s \hat j + t \hat k)$$.

To keep things general let $$x = \frac\pi6.$$ So we have: $$\cos^2x + s^2 + t^2 = 1$$ $$s^2 + t^2 = \sin^2x$$

Let's have another variable, say, $$y$$. We may write $$s$$ as $$\sin x \cos y$$ and $$t$$ as $$\sin x \sin y$$ and our equation will still hold as $$s^2 + t^2 = \sin^2x(\cos^2 y + \sin^2 y)$$ If we can find $$y$$ we can find $$s,$$ $$t$$, and hence the vector $$c$$. We have information that will help us do this. $$\overrightarrow b \cdot \overrightarrow c$$ gives us that $$\frac{\cos x}2 + \sqrt3\frac{\sin x\cos y}2 = \cos z$$ where $$z$$ is the angle between $$b$$ and $$c$$, here, $$\frac\pi4$$. So, here, we have $$\frac{\cos x}2 + \sqrt3\frac{\sin x\cos y}2 = \frac1{\sqrt2}$$ $$\cos y = \frac{2\sqrt 2}{\sqrt3} - 1$$ $$s = \sqrt{\frac 23} - \frac12$$ $$t = \sqrt{\sin^2x-s^2} = \sqrt{\sqrt\frac23-\frac23}$$ We have now the vectors required to give us the answer. Take box product $$[\overrightarrow a, \overrightarrow b, \overrightarrow c]$$. This is $$a|\overrightarrow b \times \overrightarrow c| \cos\theta$$. Divide by the magnitudes and take its $$\arccos$$.

So we have the box product as:

$$\triangle = abc\begin{vmatrix} 1&0&0\\{\frac 12}&{\frac{\sqrt3}2}&0\\\frac{\sqrt3}2&\sqrt\frac23 -1&\sqrt{\sqrt\frac23 -\frac23}\end{vmatrix}$$ And so our answer is $$\theta = \arccos\frac\triangle{abc\sin z}$$ as observed earlier ($$|b\times c| = bc\sin z$$ where $$z$$ is as we have described earlier)

Expanding $$\triangle$$ by position $$1,1$$ we get $$\boxed{\theta = \arccos \sqrt{\sqrt\frac32 -1} \approx 1.0769 rad}$$

• This was exactly my approach until the step where you substituted s and t. Very nicely done. Sep 27, 2019 at 17:00
• Could you please share the motivation behind the substitution? Sep 27, 2019 at 17:00
• Thank you! I have done a similar problem before. In it, I had taken $t$ to be $k\cdot s$. This gives $\sin^2 x = s^2(1+k^2)$. I experimented putting $k = \tan y$ and from that I was able to solve it. Later, in another book, I saw it written as I have in my answer. I feel like this is probably more instructive. (It is, of course, the same thing). Sep 27, 2019 at 17:05
• @thewitness I have fixed a silly mistake in my original answer and have provided the final answer to the question posed. Please do see it. Sep 28, 2019 at 6:13
• Only the √3/2 instead of 1/2 typo right? Sep 28, 2019 at 8:24

I just got another solution.

Consider $$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$$ to be unit vectors.

$$\left| \overrightarrow {a}\cdot \left( \overrightarrow {b}\times \overrightarrow {c}\right) \right| ^{2}=\begin{vmatrix} \overrightarrow {a}\cdot \overrightarrow {a} & \overrightarrow {a}\cdot \overrightarrow {b} & \overrightarrow {a}\cdot \overrightarrow {c} \\ \overrightarrow {b}\cdot \overrightarrow {a} & \overrightarrow {b}\cdot \overrightarrow {b} & \overrightarrow {b} \cdot \overrightarrow {c} \\ \overrightarrow {c}\cdot \overrightarrow {a} & \overrightarrow {c}\cdot \overrightarrow {b} & \overrightarrow {c} \cdot \overrightarrow{a} \end{vmatrix}$$ (product of two determinants)

$$=\begin{vmatrix} 1 & \dfrac {1}{2} & \dfrac {\sqrt {3}}{2} \\ \dfrac {1}{2} & 1 & \dfrac {1}{\sqrt {2}} \\ \dfrac {\sqrt {3}}{2} & \dfrac {1}{\sqrt {2}} & 1 \end{vmatrix}$$

$$=\frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2}$$ $$\implies \cos^2(\theta) \times \frac{1}{2} = \frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2}$$

$$\implies \theta = \cos^{-1} \left(\sqrt{\sqrt{\frac{3}{2}}-1} \right)$$ is the required angle.