Figure 2.73 Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes? Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this Chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product. The symmetric forms of two lines, L 1 and L 2 , are L 1 : x − x 1 a 1 = y − y 1 b 1 = z − z 1 c 1 L 2 : x − x 2 a 2 = y − y 2 b 2 = z − z 2 c 2 You are to develop a formula for the distance d between these two lines, in terms of the values a 1 , b 1 , c 1 ; b 2 , c 2 ; x 1 , y 1 , z 1 ; and x 2 , y 2 , z 2 . The distance between two lines is usually taken to mean the minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines. 2. Find the cross product of these two vectors and call it N . This vector is perpendicular to v 1 and v 2 , and hence is perpendicular to both lines.
Figure 2.73 Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes? Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this Chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product. The symmetric forms of two lines, L 1 and L 2 , are L 1 : x − x 1 a 1 = y − y 1 b 1 = z − z 1 c 1 L 2 : x − x 2 a 2 = y − y 2 b 2 = z − z 2 c 2 You are to develop a formula for the distance d between these two lines, in terms of the values a 1 , b 1 , c 1 ; b 2 , c 2 ; x 1 , y 1 , z 1 ; and x 2 , y 2 , z 2 . The distance between two lines is usually taken to mean the minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines. 2. Find the cross product of these two vectors and call it N . This vector is perpendicular to v 1 and v 2 , and hence is perpendicular to both lines.
Figure 2.73 Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes?
Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this Chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot
product.
The symmetric forms of two lines,
L
1
and
L
2
,
are
L
1
:
x
−
x
1
a
1
=
y
−
y
1
b
1
=
z
−
z
1
c
1
L
2
:
x
−
x
2
a
2
=
y
−
y
2
b
2
=
z
−
z
2
c
2
You are to develop a formula for the distance
d
between these two lines, in terms of the values
a
1
,
b
1
,
c
1
;
b
2
,
c
2
;
x
1
,
y
1
,
z
1
;
and
x
2
,
y
2
,
z
2
.
The distance between two lines is usually taken to mean the
minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both lines and intersects both lines.
2. Find the cross product of these two vectors and call it
N
.
This vector is perpendicular to
v
1
and
v
2
,
and hence is perpendicular to both lines.
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
Students were asked to simplify the expression (secØ - cosØ)/secØ Two students' work is given.Student A: step 1 secØ/secØ - cosØ/secØstep 2 cosØ/1 - (1/cosØ)step 3 1 - cos^2Østep 4 sin^2ØStudent B: step 1 (1/cosØ)-cosØ)/secØstep 2 (1 - cos^2Ø/cosØ)/secØstep 3 sin^2Ø/cos^2Østep 4 tan^2ØPart A: Which student simplified the expression incorrectly? Explain the errors that were made or the formulas that were misused.Part B: Complete the student's solution correctly, beginning with the location of the error.
Although 330° is a special angle on the unit circle, Amar wanted to determine its coordinates using the sum and difference formulas.Part A: Determine cos 330° using the cosine sum identity. Be sure to include all necessary work.Part B: Determine sin 330° using the sine difference identity. Be sure to include all necessary work.
A public health researcher is studying the impacts of nudge marketing techniques on shoppers vegetables
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