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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 $overrightarrow btimes overrightarrow c$

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
$$sintheta=fracoverrightarrow a times(overrightarrow b times overrightarrow c)overrightarrow a=fracb$$
$$=fracoverrightarrow b cos ac-overrightarrow c cos absin pi over 4$$
$$=fracsqrt1^2.cos^2 ac+1^2 cos^2 ab-2(cos ac)(cos ab)(cos ac)1 over sqrt2$$

Can you take it from here.

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

• $hat icdot vec b =frac12 implies b_x=frac12,; b_y=fracsqrt 32$


• $vec bcdot vec c =fracsqrt 22implies frac12 c_x+fracsqrt 32c_y=fracsqrt 22 implies c_x+sqrt 3 c_y=sqrt 2$


• $vec ccdot hat i =fracsqrt 32 implies c_x=fracsqrt 32 implies c_y=fracsqrt 2sqrt 3-frac12$

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 $b(cosfracpi 3 hat i + sinfracpi 3 hat j) = b(fracsqrt3 2 hat i + frac 12 hat j)$. We know $c$ must be of the form $c(cosfracpi6 hat i + s hat j + t hat k)=c(frac 12 hat i + s hat j + t hat k)$.

To keep things general let $x = fracpi6.$ 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 $$cos^2x + sin^2xcos y = cos z$$ where $z$ is the angle between $b$ and $c$, here, $fracpi4$.
So, here, we have $$frac 34 + frac 14cos y = frac1sqrt2 $$
$$cos y = 2sqrt 2 - 3$$.
$$s = sqrt 2 - frac32$$
$$t = sqrtsin^2x-s^2 = sqrt3sqrt2 - 4$$
We have now the vectors required to give us the answer. Take box product $[overrightarrow a, overrightarrow b, overrightarrow c]$. This is $a|overrightarrow b times overrightarrow c| costheta$. Divide by the magnitudes and take its $arccos$.