Condition for line intersection in $\mathbb{R}^3$

Solution 1:

The actual condition is

$ (\mathbf{a_2 } - \mathbf{a_1} ) \cdot (\mathbf{b_2} \times \mathbf{b_1} ) = 0 $

But since for any three vectors $\mathbf{a,b,c}$,

$ \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c} ) = \mathbf{b}\cdot (\mathbf{c} \times \mathbf{a} ) = \mathbf{c}\cdot (\mathbf{a} \times \mathbf{b} )$

then the above condition can also be written as

$ \mathbf{b_1} \cdot ( (\mathbf{a_2 } - \mathbf{a_1} ) \times \mathbf{b_2}) = 0 $

The last equation in the question can be deduced by expanding the above equation and using the cyclic equality again.

Solution 2:

Take the direction of the common normal (direction perpendicular to both lines) as

$$ \boldsymbol{n} = \boldsymbol{b}_1 \times \boldsymbol{b}_2 $$

Now project any point along each line on the common normal and it will result in a constant value

$$ d_1 = (\boldsymbol{a}_1 + \lambda_1 \boldsymbol{b}_1) \cdot \boldsymbol{n} = \boldsymbol{a}_1 \cdot \boldsymbol{n} $$

$$ d_2 =(\boldsymbol{a}_2 + \lambda_2 \boldsymbol{b}_2) \cdot \boldsymbol{n} =\boldsymbol{a}_2 \cdot \boldsymbol{n} $$

if the two lines intersect then these constants are the same $d_1 = d_2$ since they have at least one point in common.

So the condition is

$$ \boldsymbol{a}_1 \cdot \boldsymbol{n} = \boldsymbol{a}_2 \cdot \boldsymbol{n} $$

or

$$ \boldsymbol{a}_1 \cdot ( \boldsymbol{b}_1 \times \boldsymbol{b}_2) = \boldsymbol{a}_2 \cdot ( \boldsymbol{b}_1 \times \boldsymbol{b}_2) $$

or

$$ (\boldsymbol{a}_1-\boldsymbol{a}_2) \cdot (\boldsymbol{b}_1 \times \boldsymbol{b}_2) = 0 $$

Now you can rewrite the above in any other form you want as seen by the answer of Quarter Lemon or in the problem statement, using the identities of the vector triple product $a\cdot(b\times c) = b\cdot(c\times a)=c\cdot(a\times b)$.