Find the distance between the skew lines with parametric equations $ x = 1 + t , y = 1 + 6t , z = 2t $ and $ x = 1 + 2s, y = 5 + 15s , z = -2 + 6s $.
So we're giving our first line. X equals one plus t y equals one plus 60 and Z equals two t. Then we're given the equation. Where a second line or the parametric equations. So we have X equals one plus two s. Why equals five plus 15 US and Z equals negative. She plus success. We find the direction of the first line to be 162 That's based on the one. I'm sorry, the one, six and the two and then the direction of the second line Beef two is going to be 2 15, 6 and we can view the to skew lines as lying on two parallel planes. So the common normal vector is then going to be the cross product of their direction vectors. So get the normal vector to be V one. Cross speech you. When we take that cross product, we end up getting six X minus two y plus three Z plus 10 equals zero. That helps us to find that plane. So now letting T equals zero for our first line, we get a 0.100 Um, actually, it's 110 plugging this into our formula for the distance we end up getting that d is equal to six minus two plus zero plus 10 all over the magnitude of the normal vector, which is seven. So it's gonna be 14/7, which is two for our final answer.