Angle of the body diagonals

Using vector dot product calculate the angle of the body diagonals of the cube.

Result

A =  70.529 °

Solution:

D1=(1,1,1) D2=(1,1,1)  d1=D1 d1=12+12+12=31.7321  d2=D2 d2=12+12+(1)2=31.7321  D1 D2=d1 d2cosA  c=cosA=D1 D2d1d2  c=1 1+1 1+1 (1)d1 d2=1 1+1 1+1 (1)1.7321 1.7321130.3333  A0=arccos(c)=arccos(0.3333)1.231 rad  A=A0 =A0 180π  =70.52878  =70.529=703144"D_{1}=(1,1,1) \ \\ D_{2}=(1,1,-1) \ \\ \ \\ d_{1}=|D_{1}| \ \\ d_{1}=\sqrt{ 1^2+1^2+1^2 }=\sqrt{ 3 } \doteq 1.7321 \ \\ \ \\ d_{2}=|D_{2}| \ \\ d_{2}=\sqrt{ 1^2+1^2+(-1)^2 }=\sqrt{ 3 } \doteq 1.7321 \ \\ \ \\ D_{1} \cdot \ D_{2}=d_{1} \ d_{2} \cos A \ \\ \ \\ c=\cos A=\dfrac{ D_{1} \cdot \ D_{2} }{ d_{1} d_{2} } \ \\ \ \\ c=\dfrac{ 1 \cdot \ 1+1 \cdot \ 1+1 \cdot \ (-1) }{ d_{1} \cdot \ d_{2} }=\dfrac{ 1 \cdot \ 1+1 \cdot \ 1+1 \cdot \ (-1) }{ 1.7321 \cdot \ 1.7321 } \doteq \dfrac{ 1 }{ 3 } \doteq 0.3333 \ \\ \ \\ A_{0}=\arccos(c)=\arccos(0.3333) \doteq 1.231 \ \text{rad} \ \\ \ \\ A=A_{0} \rightarrow \ ^\circ =A_{0} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =70.52878 \ \ ^\circ =70.529 ^\circ =70^\circ 31'44"



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Tips to related online calculators
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator.
Pythagorean theorem is the base for the right triangle calculator.
See also our trigonometric triangle calculator.

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