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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 $.

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Wen Zheng

Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 5

Equations of Lines and Planes

Vectors

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Lectures

02:56

In mathematics, a vector (from the Latin word "vehere" meaning "to carry") is a geometric entity that has magnitude (or length) and direction. Vectors can be added to other vectors according to vector algebra. Vectors play an important role in physics, engineering, and mathematics.

11:08

In mathematics, a vector (from the Latin word "vehere" which means "to carry") is a geometric object that has a magnitude (or length) and direction. A vector can be thought of as an arrow in Euclidean space, drawn from the origin of the space to a point, and denoted by a letter. The magnitude of the vector is the distance from the origin to the point, and the direction is the angle between the direction of the vector and the axis, measured counterclockwise.

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Find the distance between …

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Show that the lines of equ…

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Find the distance between…

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.

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