   Chapter 12.5, Problem 78E

Chapter
Section
Textbook Problem

# 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.

To determine

To find: The distance between the skew lines with the parametric equations x=1+t,y=1+6t,z=2t and x=1+2s,y=5+15s,z=2+6s.

Explanation

Formula used:

The expression to find the distance between two skew lines is,

D=|nb||n| (1)

Here,

n is the normal vector, which is determined by direction vectors of two skew lines and

b is the vector joined between the points on the two skew lines.

Calculation:

The distance between two skew lines is the absolute value of the scalar projection of the vector that is joined between the points on the skew lines along the normal vector of two lines.

Write the parametric equation of the line L1 as follows.

x=1+t,y=1+6t,z=2t

Rewrite the expression as follows.

x=1+(1)t,y=1+(6)t,z=0+(2)t (2)

Write the parametric equation of the line L2 as follows.

x=1+2s,y=5+15s,z=2+6s

Rewrite the expression as follows.

x=1+(2)s,y=5+(15)s,z=2+(6)s (3)

The expressions for the parametric equations for a line through the point (x0,y0,z0) and parallel to the direction vector a,b,c is,

x=x0+at,y=y0+bt,z=z0+ct (4)

Compare equation (2) with equation (4) and write the direction vector of line L1 as follows.

v1=1,6,2

Compare equation (3) with equation (4) and write the direction vector of line L2 as follows.

v2=2,15,6

The direction vectors of two skew lines are written as follows.

v1=1,6,2v2=2,15,6

The normal vector is the cross product of direction vectors of both the lines.

Write the expression to find normal vector as follows.

n=|ijka1b1c1a2b2c2| (5)

Substitute 1 for a1, 6 for b1, 2 for c1, 2 for a2, 15 for b2, and 6 for c2 in equation (5),

n=|ijk1622156|=i[(6)(6)(15)(2)]j[(1)(6)(2)(2)]+k[(1)(15)(2)(6)]=ij+k=6i2j+3k

Consider the scalar parameters t and s as 0 and obtain the two points on the two skew lines as follows.

Substitute 0 for t in equation (2) and obtain a point on the line L1.

x=1+(0),y=1+6(0),z=2(0)x=1,y=1,z=0

The point on the line L1 is (1,1,0).

Substitute 0 for s in equation (3) and obtain a point on the line L2.

x=1+2(0),y=5+15(0),z=2+6(0)x=1,y=5,z=2

The point on the line L2 is (1,5,2)

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