Three angles of a triangle ABC in R^3 (not as hard as it seems)

I cant find the angle between the vertices c (1,1,1) a (1,-4,2) b (-5,2,7) Can you help me?

I set:

$\small{ \begin{array}{lcr} a_x &=& 1\\ a_y&=&-4\\ a_z&=&2 \end{array} \qquad \begin{array}{lcr} b_x &=& -5\\ b_y&=&2\\ b_z&=&7 \end{array} \qquad \begin{array}{lcr} c_x &=& 1\\ c_y &=& 1\\ c_z &=& 1\\ \end{array} }$

Differences of the vectors:
$\small{ \begin{array}{lcr} \vec{b}-\vec{a} &=& \bigl(\begin{smallmatrix} b_x-a_x\\b_y-a_y\\b_z-a_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr)\\ \vec{a}-\vec{b} &=& -\bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr) \\ \end{array} \qquad \begin{array}{lcr} \vec{c}-\vec{b} &=& \bigl(\begin{smallmatrix} c_x-b_x\\c_y-b_y\\c_z-b_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr)\\ \vec{b}-\vec{c} &=& -\bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr) \\ \end{array} \qquad \begin{array}{lcr} \vec{a}-\vec{c} &=& \bigl(\begin{smallmatrix} a_x-c_x\\a_y-c_y\\a_z-c_z \end{smallmatrix}\bigr) = \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr)\\ \vec{c}-\vec{a} &=& -\bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr) \\ \end{array} }$

In the plane of the triangle (ABC) the angles are:

$\small{ \begin{array}{rcl} \tan{(A)} &=& \frac{ \vert (\vec{c}-\vec{a}) \times (\vec{b}-\vec{a}) \vert} {(\vec{c}-\vec{a})\cdot (\vec{b}-\vec{a}) } = \frac{ \left| \bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} 0\\5\\-1 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} -6\\6\\5 \end{smallmatrix}\bigr) } = \frac{ 43.5545634808 } {25 } \qquad (I.)\\ A &=& \arctan{(1.74218253923)}\\ A &=& 60.1444921468^\circ \\ \tan{(B)} &=& \frac{ \vert (\vec{a}-\vec{b}) \times (\vec{c}-\vec{b}) \vert} {(\vec{a}-\vec{b})\cdot (\vec{c}-\vec{b}) } = \frac{ \left| \bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} 6\\-6\\-5 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} 6\\-1\\-6 \end{smallmatrix}\bigr) } = \frac{ 43.5545634808 } {72 } \qquad (I.)\\ B &=& \arctan{(0.60492449279)}\\ B &=& 31.1707710617^\circ \\ \tan{(C)} &=& \frac{ \vert (\vec{b}-\vec{c}) \times (\vec{a}-\vec{c}) \vert} {(\vec{b}-\vec{c})\cdot (\vec{a}-\vec{c}) } = \frac{ \left| \bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr) \times \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr) \right|} {\bigl(\begin{smallmatrix} -6\\1\\6 \end{smallmatrix}\bigr)\cdot \bigl(\begin{smallmatrix} 0\\-5\\1 \end{smallmatrix}\bigr)} = \frac{ 43.5545634808 } {1 } \qquad (I.)\\ C &=& \arctan{(43.5545634808)}\\ C &=& 88.6847367915^\circ \\ \end{array} }$

$A+B+C = 60.1444921468^\circ + 31.1707710617^\circ + 88.6847367915^\circ = 180^\circ$

Thanks, heureka......here's another method that is vey similar to its 2D counterpart....

c (1,1,1) a (1,-4,2) b (-5,2,7)

Let vector ca =  < 0, -5, 1>

Let vector cb =  < -6, 1, 6 >

Find the length of ca =  sqrt26]

Find the length of cb =sqrt [73]

Find the dot product of both vectors = [0*-6 + -5*1 + 1*6]   = [1]

And the measure of angle acb can be found as follows :

cos[acb] = dot product of both vectors / [ product of their lengths]   .......so we have

cos[acb]  = 1 / sqrt[73*26]

cos^-1 [ 1/sqrt[73*26]  = acb ≈ 88.684736791499°

Similarly 

Let vector ba = < 6,-6,-5>     and its length = sqrt(97)

Let vector bc = < 6, -1, -6>     and it's length = sqrt(73)

And the dot probuct of both vectors = [36 + 6 + 30]  =[72]

So

cos(abc)  = 72/ sqrt(97*73)

cos^-1 [ 72/ sqrt(97*73)]  = abc  ≈ 31.17077106171°

Lastly

Let vector ab  = < -6, 6, 5>     and its length = sqrt(97)

Let vector ac = <0, 5, -1>     and it's length  = sqrt(26)

And the dot product of the two vectors = [0 + 30 - 5 ]  = [25]

So

cos(bac)  = 25 / sqrt(97*26) 

cos^-1 [ 25 / sqrt(97*26) ] = bac ≈ 60.144492146791°

Just as heureka found......

Yet another approach is simply to use Pythagoras to find the length of each side, then use the cosine rule to find the angles:

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