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$$ \begin{align} \color{#C00000}{((10,10,6)-(5,5,4))}\times\color{#00A000}{((10,10,3)-(5,5,5))} &=\color{#C00000}{(5,5,2)}\times\color{#00A000}{(5,5,-2)}\\ &=\color{#0000FF}{(-20,20,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(-20,20,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $0$, so the lines intersect. $$ \color{#C00000}{(5,5,2)}\times\color{#0000FF}{(-20,20,0)}=\color{#E06800}{(-40,-40,200)} $$ is perpendicular to a plane containing the first line; therefore, $\color{#E06800}{(-40,-40,200)}\cdot u$ is constant along the first line. In this case, that constant is $400$. The general point along the second line is $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(-40,-40,200)}\cdot(\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t)=400\\ 600-800t=400\\ t=1/4 $$ Plugging $t=1/4$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}\cdot1/4=\left(\frac{25}{4},\frac{25}{4},\frac{9}{2}\right) $$

The second example $$ \begin{align} \color{#C00000}{((12,15,4)-(6,8,4))}\times\color{#00A000}{((12,15,6)-(6,8,2))} &=\color{#C00000}{(6,7,0)}\times\color{#00A000}{(6,7,4)}\\ &=\color{#0000FF}{(28,-24,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(28,-24,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $-24$, so the lines intersect. $$ \color{#C00000}{(6,7,0)}\times\color{#0000FF}{(28,-24,0)}=\color{#E06800}{(0,0,-340)} $$ is perpendicular to a plane containing the first line; therefore, $\color{#E06800}{(0,0,-340)}\cdot u$ is constant along the first line. In this case, that constant is $-1360$. The general point along the second line is $$ \color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(0,0,-340)}\cdot(\color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t)=-1360\\ -680-1360t=-1360\\ t=1/2 $$ Plugging $t=1/2$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}\cdot1/2=\left(9,\frac{23}{2},4\right) $$

$$ \begin{align} \color{#C00000}{((10,10,6)-(5,5,4))}\times\color{#00A000}{((10,10,3)-(5,5,5))} &=\color{#C00000}{(5,5,2)}\times\color{#00A000}{(5,5,-2)}\\ &=\color{#0000FF}{(-20,20,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(-20,20,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $0$, so the lines intersect. $$ \color{#C00000}{(5,5,2)}\times\color{#0000FF}{(-20,20,0)}=\color{#E06800}{(-40,-40,200)} $$ is perpendicular to the first line; therefore, $\color{#E06800}{(-40,-40,200)}\cdot u$ is constant along the first line. In this case, that constant is $400$. The general point along the second line is $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(-40,-40,200)}\cdot(\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t)=400\\ 600-800t=400\\ t=1/4 $$ Plugging $t=1/4$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}\cdot1/4=\left(\frac{25}{4},\frac{25}{4},\frac{9}{2}\right) $$

The second example $$ \begin{align} \color{#C00000}{((12,15,4)-(6,8,4))}\times\color{#00A000}{((12,15,6)-(6,8,2))} &=\color{#C00000}{(6,7,0)}\times\color{#00A000}{(6,7,4)}\\ &=\color{#0000FF}{(28,-24,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(28,-24,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $-24$, so the lines intersect. $$ \color{#C00000}{(6,7,0)}\times\color{#0000FF}{(28,-24,0)}=\color{#E06800}{(0,0,-340)} $$ is perpendicular to the first line; therefore, $\color{#E06800}{(0,0,-340)}\cdot u$ is constant along the first line. In this case, that constant is $-1360$. The general point along the second line is $$ \color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(0,0,-340)}\cdot(\color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t)=-1360\\ -680-1360t=-1360\\ t=1/2 $$ Plugging $t=1/2$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}\cdot1/2=\left(9,\frac{23}{2},4\right) $$

$$ \begin{align} \color{#C00000}{((10,10,6)-(5,5,4))}\times\color{#00A000}{((10,10,3)-(5,5,5))} &=\color{#C00000}{(5,5,2)}\times\color{#00A000}{(5,5,-2)}\\ &=\color{#0000FF}{(-20,20,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(-20,20,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $0$, so the lines intersect. $$ \color{#C00000}{(5,5,2)}\times\color{#0000FF}{(-20,20,0)}=\color{#E06800}{(-40,-40,200)} $$ is perpendicular to a plane containing the first line; therefore, $\color{#E06800}{(-40,-40,200)}\cdot u$ is constant along the first line. In this case, that constant is $400$. The general point along the second line is $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(-40,-40,200)}\cdot(\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t)=400\\ 600-800t=400\\ t=1/4 $$ Plugging $t=1/4$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}\cdot1/4=\left(\frac{25}{4},\frac{25}{4},\frac{9}{2}\right) $$

$$ \begin{align} \color{#C00000}{((10,10,6)-(5,5,4))}\times\color{#00A000}{((10,10,3)-(5,5,5))} &=\color{#C00000}{(5,5,2)}\times\color{#00A000}{(5,5,-2)}\\ &=\color{#0000FF}{(-20,20,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(-20,20,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $0$, so the lines intersect. $$ \color{#C00000}{(5,5,2)}\times\color{#0000FF}{(-20,20,0)}=\color{#E06800}{(-40,-40,200)} $$ is perpendicular to a plane containing the first line; therefore, $\color{#E06800}{(-40,-40,200)}\cdot u$ is constant along the first line. In this case, that constant is $400$. The general point along the second line is $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(-40,-40,200)}\cdot(\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t)=400\\ 600-800t=400\\ t=1/4 $$ Plugging $t=1/4$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}\cdot1/4=\left(\frac{25}{4},\frac{25}{4},\frac{9}{2}\right) $$

The second example $$ \begin{align} \color{#C00000}{((12,15,4)-(6,8,4))}\times\color{#00A000}{((12,15,6)-(6,8,2))} &=\color{#C00000}{(6,7,0)}\times\color{#00A000}{(6,7,4)}\\ &=\color{#0000FF}{(28,-24,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(28,-24,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $-24$, so the lines intersect. $$ \color{#C00000}{(6,7,0)}\times\color{#0000FF}{(28,-24,0)}=\color{#E06800}{(0,0,-340)} $$ is perpendicular to a plane containing the first line; therefore, $\color{#E06800}{(0,0,-340)}\cdot u$ is constant along the first line. In this case, that constant is $-1360$. The general point along the second line is $$ \color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(0,0,-340)}\cdot(\color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}t)=-1360\\ -680-1360t=-1360\\ t=1/2 $$ Plugging $t=1/2$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(6,8,2)}+\color{#00A000}{(6,7,4)}\cdot1/2=\left(9,\frac{23}{2},4\right) $$

$$ \begin{align} \color{#C00000}{((10,10,6)-(5,5,4))}\times\color{#00A000}{((10,10,3)-(5,5,5))} &=\color{#C00000}{(5,5,2)}\times\color{#00A000}{(5,5,-2)}\\ &=\color{#0000FF}{(-20,20,0)} \end{align} $$ is perpendicular to both lines; therefore, $(-20,20,0)\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $0$, so the lines intersect. $$ \color{#C00000}{(5,5,2)}\times\color{#0000FF}{(-20,20,0)}=(-40,-40,200) $$ is perpendicular to a plane containing the first line; therefore, $(-40,-40,200)\cdot u$ is constant along the first line. In this case, that constant is $400$. The general point along the second line is $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t $$ To compute the point of intersection, find the $t$ so that $$ (\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t)\cdot(-40,-40,200)=400\\ 600-800t=400\\ t=1/4 $$ Plugging $t=1/4$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}\cdot1/4=\left(\frac{25}{4},\frac{25}{4},\frac{9}{2}\right) $$

$$ \begin{align} \color{#C00000}{((10,10,6)-(5,5,4))}\times\color{#00A000}{((10,10,3)-(5,5,5))} &=\color{#C00000}{(5,5,2)}\times\color{#00A000}{(5,5,-2)}\\ &=\color{#0000FF}{(-20,20,0)} \end{align} $$ is perpendicular to both lines; therefore, $\color{#0000FF}{(-20,20,0)}\cdot u$ is constant along each line. If this constant is not the same for each line, the lines do not intersect. In this case, the constant for each line is $0$, so the lines intersect. $$ \color{#C00000}{(5,5,2)}\times\color{#0000FF}{(-20,20,0)}=\color{#E06800}{(-40,-40,200)} $$ is perpendicular to a plane containing the first line; therefore, $\color{#E06800}{(-40,-40,200)}\cdot u$ is constant along the first line. In this case, that constant is $400$. The general point along the second line is $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t $$ To compute the point of intersection, find the $t$ so that $$ \color{#E06800}{(-40,-40,200)}\cdot(\color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}t)=400\\ 600-800t=400\\ t=1/4 $$ Plugging $t=1/4$ into the formula for a point along the second line, yields the point of intersection: $$ \color{#00A000}{(5,5,5)}+\color{#00A000}{(5,5,-2)}\cdot1/4=\left(\frac{25}{4},\frac{25}{4},\frac{9}{2}\right) $$