Angle between a vector and cross product of two vectors

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

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 $$\boxed{\theta = \arccos \sqrt{\sqrt\frac32 -1} \approx 1.0769 rad}$$