Q – consider a cube in $\mathbb R^3$ with unit side length and one vertex at the origin. Find the angle between the space diagonal of this cube and one of its edges.
my solution/attempt is attached, please correct if needed.
Find angle between space diagonal and one of its edges of cube with unit side length.
geometryvectors
Related Solutions
In the picture shown, $AB = DC = 10$, $BC = AD = 6$, and $\angle BDC = 30^{\circ}$.
Apply the Law of Cosines to $\triangle BDC$. Then
$$6^2 = d^2 + 10^2 - 2 \cdot d \cdot 10 \cdot \cos 30^{\circ}$$ $$36 = d^2 + 100 - 20d \cdot \frac{\sqrt{3}}{2}$$ $$0 = d^2 - 10d\sqrt{3} + 64$$ $$d^2=10d\sqrt{3} + 64$$ $$d=\sqrt{10d\sqrt{3} + 64}$$
Apply the quadratic formula to solve for the long diagonal $d$. You should get two (geometrically valid) roots, although only one of the roots makes $BD$ the long diagonal; the other root makes it the short one.
The angle between a segment and the OX axis given segment origin, length and a perpendicular segment
You do not have enough information. take a sheet of paper, put it on a table such that one corner $A$ is fixed, and the paper can rotate around $A$. Can you tell me what the angle is between the paper and the edge of the table? I have $A$, $L_1$ and $L_2$, and those are perpendicular, but the angle can take any value since I can rotate the paper.
Best Answer
The answer is right. Alternatively, since you marked it with the
vectorstag, you might want to use dot product. Starting from origin, the diagonal is $(1,1,1)$. The side is $(1,0,0)$ (or $(0,1,0)$ or $(0,0,1)$). Then $$(1,1,1)\cdot(1,0,0)=1=|(1,1,1)|\cdot |(1,0,0)|\cdot\cos\theta=\sqrt3 \cdot 1\cdot\cos\theta$$ Therefore $$\theta=\arccos\frac1{\sqrt 3}$$ This is the same angle.