Example 2.6. Show that the following lines are skew (nonparallel and nonintersecting) and find the distance between them. x = 1+7t, y = 3+t, z = 5 – 3t and x = 4 – s, y = 6, z = 7+ 2s. The direction vectors of the two lines are u =< 7, 1, –3 > and v =< -1,0,2 > . If they are skew they must lie in parallel planes, the common normal to which is n = u x v =< 2, –11,1 > . The shortest distance between the two lines is the distance along the common normal. Consider the vector joining two arbitrary points, one on the first line, B(1,3, 5), and another on the second line, D(4, 6, 7) given by w = BD =< 3, 3, 2 > . The required distance D= PQ is the component of w along n, i.e., < 3, 3, 2 > ·< 2, –11, 1 >| |< 2, –11, 1 > I 25 D = w - |n| V126

Linear Algebra: A Modern Introduction
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ISBN:9781285463247
Author:David Poole
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Chapter4: Eigenvalues And Eigenvectors
Section4.2: Determinants
Problem 10AEXP
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why n is<2,-11,1>

Example 2.6. Show that the following lines are skew (nonparallel and nonintersecting) and find the
distance between them.
x = 1+7t, y = 3 +t, z = 5 – 3t and x = 4 – s, y = 6, z = 7+2s.
The direction vectors of the two lines are u =< 7, 1, –3 > and v =< -1,0,2 > . If they are skew
they must lie in parallel planes, the common normal to which is
n = u x v =< 2, –11,1 > .
The shortest distance between the two lines is the distance along the common normal. Consider the
vector joining two arbitrary points, one on the first line, B(1,3, 5), and another on the second line,
D(4,6, 7) given by w = BD =< 3,3,2 >. The required distance D = PQ is the component of w
along n, i.e.,
< 3, 3, 2 > · < 2, –11, 1 >
|< 2, –11,1 > |
25
n
D = w ·
n
V126
Transcribed Image Text:Example 2.6. Show that the following lines are skew (nonparallel and nonintersecting) and find the distance between them. x = 1+7t, y = 3 +t, z = 5 – 3t and x = 4 – s, y = 6, z = 7+2s. The direction vectors of the two lines are u =< 7, 1, –3 > and v =< -1,0,2 > . If they are skew they must lie in parallel planes, the common normal to which is n = u x v =< 2, –11,1 > . The shortest distance between the two lines is the distance along the common normal. Consider the vector joining two arbitrary points, one on the first line, B(1,3, 5), and another on the second line, D(4,6, 7) given by w = BD =< 3,3,2 >. The required distance D = PQ is the component of w along n, i.e., < 3, 3, 2 > · < 2, –11, 1 > |< 2, –11,1 > | 25 n D = w · n V126
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