Question

Using the relationship between the direction cosines of perpendicular lines in space, prove that, for lines l₁ and l₂, with the direction angles α₁, β₁, γ₁, and α₂, β₂, γ₂, cosα₁cosα₂ + cosβ₁cosβ₂ + cosγ₁cosγ₂ = 0 How could this be proven?

Answer 1

Using the definitions of direction cosines:

For the line `l_1`, we can denote the direction vector as:

`bb(ul u) = u_1bb(ul hat i) + u_2bb(ul hat j) + u_3bb(ul hat j)`

Where:

`cos alpha_1 = u_1 / || bb(ul u) ||`, `cos beta_1 = u_2 / || bb(ul u) ||`, `cos gamma_1 = u_3 / || bb(ul u) ||`

Similarly, For the line `l_2`, we can denote the direction vector as:

`bb(ul v) = v_1bb(ul hat i) + v_2bb(ul hat j) + v_3bb(ul hat j)`

Where:

`cos alpha_2 = v_1 / || bb(ul v) ||`, `cos beta_2 = v_2 / || bb(ul v) ||`, `cos gamma_2 = v_3 / || bb(ul v) ||`

Given this we denote the expression we seek by:

`E = cos alpha_1 cos alpha_2 + cos beta_1 cos beta_2 + cos gamma_1 cos gamma_2`

`\ \ \ = u_1 / || bb(ul u) || v_1 / || bb(ul v) || + u_2 / || bb(ul u) || v_2 / || bb(ul v) || + u_3 / || bb(ul u) || v_3 / || bb(ul v) ||`

`\ \ \ = (u_1 v_1 + u_2 v_2 + u_3 v_3) / ( ||bb(ul u)|| \ ||bb(ul v)|| )`

`\ \ \ = (bb(ul u) * bb(ul v)) / ( ||bb(ul u)|| \ ||bb(ul v)|| )`

Now, we can use the fact that if `bb(ul u)` and `bb(ul v)` are perpendicular then the dot (or scalar) product is zero, ie

`bb(ul u) * bb(ul v) = 0 => E = (0) / ( ||bb(ul u)|| \ ||bb(ul v)|| )`

And hence we have the desired result:

`cos alpha_1 cos alpha_2 + cos beta_1 cos beta_2 + cos gamma_1 cos gamma_2 = 0 \ \ \ \` QED