How to find an angle between two sides of cube?

Question

Given $\text{Cube}\ ABCDA_1B_1C_1D_1$ Find the angle between $AB_1$ and $BD_1$ Usually I will find a parallel line which has common point with the other line, however I can't find such line in this exercise. Here is a picture of a cube, the red lines are the one which I need to find the angle between. Also the answer in my textbook is 60 degrees.

Answer 1

You don't need vectors. Just a little trick and Pythagoras Theorem.

Assume WLOG that it's a unit cube. Draw an identical second cube just below the first and translate $AB_1$ to the corresponding vertices below. The translation preserves the angle. Let's label the vertices of the "mirror cube" with the prime symbol, e.g. $A'B_1'$, where $B_1'$ is coincident with $B$ in the original cube.

Now consider the triangle $A'BD_1$. Find the sides using Pythagoras theorem:

$A'B = \sqrt 2, BD_1 = \sqrt 3, A'D_1 = \sqrt 5$. Hence prove it's a right triangle, and you have your required angle.

Answer 2

Find angle by taking vector dot (scalar) product of the vectors of the lines/ sides of the cube.

$$ cos^{-1} {(AB_1\cdot BD_1)/(|AB_1||BD_1|)} $$

whether or not the lines intersect.

EDIT 1:

Let $ AB,BC, AA_1 $ be Cartesian coordinate axes with unit vectors $ i,j,k $ along positive direction in that sense,

$$ BD_1 = (i,-j,-k) ; AB_1 = (i,0,k) ; $$

Their cross product = +1 + 0 -1 = 0 , i.e., they are perpendicular, they make $90^0$ with each other.

[ To find shortest distance you can get the two points distant $ \delta $ between them ( along common perpendicular by cross product direction of unit vectors along them), by vector shift through $\delta$. But all this not necessary and advanced at this stage, I mention so you may be comfortable thinking in a plane geometry like situation].

Answer 3

Hint consider it to be a unit cube. With origin at A and $i,j,k$ are unit vectors in $x,y,z$ axis respectively so points are $A(0,0,0),B(1,0,0),B1(1,1,0),D1(0,1,1)$ now with vectors form equations of AB1,D1B using triangle law of vectors then using dot product $\frac{A.B}{|A|.|B|}=cos(\theta)$ and get the required angle.

Answer 4

let $a$ be the side of cube. Consider the vertex $D$ at the origin and $DA$, $DC$ & $DD_1$ coinciding with $x$, $y$ & $z$ axes respectively. $A(a, 0, 0)$, $B(a, a, 0)$, $C(0, a, 0)$, $D(0, 0, a)$, $A_1(a, 0, a)$, $B_1(a, a, a)$, $C_1(0, a, a)$ & $D_1(0, 0, a)$

By using vector, one should get $$\vec{AB_1}=a(j+k)$$ & $\vec{BD_1}=a(-i-j+k)$ hence, the angle between $\vec{AB_1}$ & $\vec{BD_1}$ is given $$\cos\theta=\frac{\vec{AB_1}\cdot \vec{BD_1}}{|\vec{AB_1}||\vec{BD_1}|}$$ $$=\frac{a(j+k)\cdot a(-i-j+k)}{|a(j+k)||a(-i-j+k)|}$$ $$=\frac{-1+1}{\sqrt 2\sqrt 3}$$ $$=\frac{0}{\sqrt 2\sqrt 3}=0$$ $$\theta=\cos^{-1}(0)=\color{red}{90^\circ}$$