Please help with #6 through #9. Thank you hence is perpendicular to both lines.3. From vector N, form a unit vector n in the same direction.4. Use symmetric equations to find a convenient vector v12 that lies between any two points, one on each line.Again, this can be done directly from the symmetric equations.5. The dot product of two vectors is the magnitude of the projection of one vector onto the other-that is,A B = || A || || B || cos 0, where is the angle between the vectors. Using the dot product, find theprojection of vector v12 found in step 4 onto unit vector n found in step 3. This projection is perpendicularto both lines, and hence its length must be the perpendicular distance d between them. Note that the value ofd may be negative, depending on your choice of vector v12 or the order of the cross product, so use absolutevalue signs around the numerator.6. Check that your formula gives the correct distance of |-25/V198 1.78 between the following two lines:7. Is your general expression valid when the lines are parallel? If not, why not? (Hint: What do you know aboutthe value of the cross product of two parallel vectors? Where would that result show up in your expression ford?)8. Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersectif they are not parallel and they are in the same plane. Hence, consider the direction of n and v12. What isthe result of their dot product?Consider the following application. Engineers at a refinery have determined they need to install support strutsbetween many of the gas pipes to reduce damaging vibrations. To minimize cost, they plan to install these strutsat the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure,they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute themto the installation crews without spending valuable time making measurements.The rectangular frame structure has the dimensions 4.0 x 15.0 x 10.0 m (height, width, and depth). Onesector has a pipe entering the lower corner of the standard frame unit and exiting at the diametrically opposedcormer (the one farthest away at the top); call this L1. A second pipe enters and exits at the two different9.opposite lower corners; call this L2 (Figure 2.74).4 m15 m10 ma standard frame unit. Distance between Two Skew LinesFigure 2.73 Industrial pipe installations often feature pipes running indifferent 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 thelines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirmingthat the distance between them meets specifications can be both important and awkward to measure. One way is tomodel the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. Thecalculation involves forming vectors along the directions of the lines and using both the cross product and the dotproduct.The symmetric forms of two lines, L, and L2, areL :b1y-y2b2You are to develop a formula for the distance d between these two lines, in terms of the valuesa1, b1, C1;a2, b2, c2; x1, y1, z1; and r2, y2, 22. The distance between two lines is usually taken to mean theminimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both linesand intersects both lines.1. First, write down two vectors, v and v, that lie along L, and L2, respectively.2. Find the cross product of these two vectors and call it N. This vector is perpendicular to V and v2, and1

Name: Please help with #6 through #9. Thank you hence is perpendicular to bo
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Description: Please help with #6 through #9. Thank you hence is perpendicular to both lines.3. From vector N, form a unit vector n in the same direction.4. Use symmetric equations to find a convenient vector v12 that lies between any two points, one on each line.

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sign 6. Check that your formula gives the correct distance of |-25/V198 1.78 between the following two lines: at de uGu knGL abo