document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1121153>Question 1121153</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. (I haven’t learned any of this, but I am trying to become familiar with it.) I found a website full of problems of the content we will be learning, but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:\r<BR>\n" );
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document.write( "I’m going to do my best to talk through each part and tell you how I think it should be solved!\r<BR>\n" );
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document.write( "A line L1 passes through points P(&#8722;1,6,&#8722;1) and Q(0,4,1). <BR>\n" );
document.write( "a. (i) Show that vector PQ=(1,&#8722;2,2).<BR>\n" );
document.write( "So, I can see that if you take the x, y, and z of Q, and subtract the x, y and z of P, then you get PQ. But, is there some sort of formula for this that you would use to answer this question?<BR>\n" );
document.write( "(ii) Hence, write down an equation for L1 in the form r=a+tb.<BR>\n" );
document.write( "Would this be: r=(&#8722;1,6,&#8722;1)+t(0,4,1)? (I just plugged in P for a and Q for b.  <BR>\n" );
document.write( "b. A second line L2 has equation r=(4,2,&#8722;1)+s(3,0,&#8722;4). <BR>\n" );
document.write( "Find the cosine between vector PQ and L2. <BR>\n" );
document.write( "I’m super confused on how you could find the cosine of this when all you have is some points? Or, if I draw this out somehow would I find the sides to do the Law of Cosines?\r<BR>\n" );
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document.write( "c. The lines L1 and L2  intersect at the point R. Find the coordinates of R. <BR>\n" );
document.write( "Would I use a midpoint formula to discover this? Since R would be in the middle of both lines? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #736947 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=736947\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A line L1 passes through points P(&#8722;1,6,&#8722;1) and Q(0,4,1). \r<BR>\n" );
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document.write( "a. (i) Show that vector PQ=(1,&#8722;2,2).\r<BR>\n" );
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document.write( "vector PQ is created from line segment PQ.\r<BR>\n" );
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document.write( "a vector has both magnitude and direction.\r<BR>\n" );
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document.write( "vector PQ = <x2-x1,y2-y1,z2-z1> = <0--1,4-6,1--1> = <1,-2,2>\r<BR>\n" );
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document.write( "note that -- means plus.<BR>\n" );
document.write( "you are subtracting a negative number, which is the same as adding a positive number.\r<BR>\n" );
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document.write( "it has direction because from, any point in space, it can only go 1 unit in the direction of the x-axis, -2 units in the direction of the y-axis, and 2 units in the direction of the z-axis.\r<BR>\n" );
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document.write( "the length of vector V is sqrt(1^2 + (-2)^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9).\r<BR>\n" );
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document.write( "(ii) Hence, write down an equation for L1 in the form r=a+tb.<BR>\n" );
document.write( "Would this be: r=(&#8722;1,6,&#8722;1)+t(0,4,1)? (I just plugged in P for a and Q for b.\r<BR>\n" );
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document.write( "the vector equation of a line is in the form of R = R0 + T * vector V.\r<BR>\n" );
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document.write( "R is any point on the line which is the same as any position vector that terminates on the line.\r<BR>\n" );
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document.write( "R0 is a specific point on the line which is the same as a a position vector that terminates on a specific point on the line.\r<BR>\n" );
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document.write( "V is a direction Vector.\r<BR>\n" );
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document.write( "the point P from the line PQ serves as the point R0 which is the same as position vector R0.\r<BR>\n" );
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document.write( "note that any point on the graph also serves as a position vector.<BR>\n" );
document.write( "a position vector is a vector that has its tail at the origin and its head terminating at the point.\r<BR>\n" );
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document.write( "therefore point (-1,6,-1) is also position vector <-1,6,-1).\r<BR>\n" );
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document.write( "this means that vector arithmetic applies to it just as with any other vector.\r<BR>\n" );
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document.write( "vector V is the same as vector PQ that we just created.\r<BR>\n" );
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document.write( "therefore, in the equation R = R0 + T * vector V, we get:\r<BR>\n" );
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document.write( "vector R0 = <-1,6,-1><BR>\n" );
document.write( "vector V = <1,-2,2>\r<BR>\n" );
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document.write( "vector R = <-1,6,-1> + T * <1,-2,2> which becomes:\r<BR>\n" );
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document.write( "vector R = <-1,6,-1> + <T,-2T,2T> which becomes:\r<BR>\n" );
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document.write( "vector R = <-1+T,6-2T,-1+2T>, where T is any real number.\r<BR>\n" );
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document.write( "that's the equation of the line R in vector equation form.\r<BR>\n" );
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document.write( "x = -1 + T<BR>\n" );
document.write( "y = 6 - 2T<BR>\n" );
document.write( "z = -1 + 2T\r<BR>\n" );
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document.write( "b. A second line L2 has equation r=(4,2,&#8722;1)+s(3,0,&#8722;4). \r<BR>\n" );
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document.write( "the equation of the first line is R = (-1,6,-1) + T * <1,-2,2><BR>\n" );
document.write( "the equation of the second line is L1 = (4,2,-1) + S * <3,0,-4>\r<BR>\n" );
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document.write( "the vector portion of these two lines is <1,-2,2> and <3,0,-4>\r<BR>\n" );
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document.write( "we'll call <1,-2,2> vector V and we'll call <3,0,-4> vector W.\r<BR>\n" );
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document.write( "these two vectors can be positioned anywhere on the graph, so they can be brought to intersect at a common point, allowing us to find the angle between them.\r<BR>\n" );
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document.write( "for example, you can place the tail end of both vectors at the origin.\r<BR>\n" );
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document.write( "the dot product of these two vectors can be found by the formula \r<BR>\n" );
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document.write( "vector V dot vector W = vx*wx + vy*wy + vz*wz.\r<BR>\n" );
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document.write( "that makes vector V dot vector W equal to 1*3 - 2*0 - 2*4 = -5.\r<BR>\n" );
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document.write( "there is an alternate formula for the dot product.\r<BR>\n" );
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document.write( "it is vector V dot vector W = magnitude of vector V * magnitude of vector W * cosine (theta), where theta is the angle between the two vectors.\r<BR>\n" );
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document.write( "you can use this formula to solve for the angle between the two vectors.\r<BR>\n" );
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document.write( "you get cosine (theta) = (vector V dot vector W) / (magnitude of vector V * magnitude of vector W).\r<BR>\n" );
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document.write( "we already calculated the dot product of vector V and vector W, which is equal to -5.\r<BR>\n" );
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document.write( "the magnitude of vector V is equal to sqrt(1^2 + (-2)^2 + 2^2) = sqrt(9)\r<BR>\n" );
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document.write( "the magnitude of vector W is equal to sqrt(3^2 + 0^2 + (-4)^2) = sqrt(25) = 5\r<BR>\n" );
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document.write( "cosine (theta) is therefore equal to -5 / (5*sqrt(9)) which is equal to -1/3.\r<BR>\n" );
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document.write( "in the first quadrant, the angle would be arcosine (1/3) = 70.52877936.\r<BR>\n" );
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document.write( "since the cosine is negative, the angle could be in the second quadrant or the third quadrant.\r<BR>\n" );
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document.write( "in the second quadrant, the angle would be (180 - 70.52877936) = 109.4712206.\r<BR>\n" );
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document.write( "in the third quadrant, the angle would be (180 + 70.52877936) = 250.5287794.\r<BR>\n" );
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document.write( "it's difficult to see what the angle is on the 3 dimension graph because the angle between the two vectors is in a different plane.\r<BR>\n" );
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document.write( "my guess is that the angle is 109.4712206 because 250 degrees just looks too large.\r<BR>\n" );
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document.write( "i used an online calculator and it told me the angle was 109 degrees, so i think my guess is right, assuming the calculator did the job correctly.\r<BR>\n" );
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document.write( "that calculator can be found at <a  rel=nofollow HREF= \"https://www.emathhelp.net/calculators/linear-algebra/angle-between-two-vectors-calculator/?ux=1&uy=-2&uz=2&vx=3&vy=0&vz=-4&steps=on\" target = \"_blank\">https://www.emathhelp.net/calculators/linear-algebra/angle-between-two-vectors-calculator/?ux=1&uy=-2&uz=2&vx=3&vy=0&vz=-4&steps=on</a>\r<BR>\n" );
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document.write( "c. The lines L1 and L2 intersect at the point R. Find the coordinates of R. <BR>\n" );
document.write( "Would I use a midpoint formula to discover this? Since R would be in the middle of both lines?\r<BR>\n" );
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document.write( "the intersection of the lines is the common point between the lines.\r<BR>\n" );
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document.write( "if you have a 3D graphing calculator, you might have the calculator find the point for you.\r<BR>\n" );
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document.write( "if not, then you need to find it algebraically by solving the two equations of the lines simultaneously.\r<BR>\n" );
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document.write( "a good reference on how to do that can be found at <a  rel=nofollow HREF= \"https://www.youtube.com/watch?v=LxDM5-wRAzI\" target = \"_blank\">https://www.youtube.com/watch?v=LxDM5-wRAzI</a>\r<BR>\n" );
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document.write( "the vector equations of both lines L1 and L2 would be:\r<BR>\n" );
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document.write( "L1 = <-1+T,6-2T,-1+2T>\r<BR>\n" );
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document.write( "L2 = <4+3S,2,-1-4S>\r<BR>\n" );
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document.write( "this means:\r<BR>\n" );
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document.write( "variable            L1                L2\r\n" );
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document.write( "x                   -1 + T            4 + 3S\r\n" );
document.write( "y                    6 - 2T           2\r\n" );
document.write( "z                   -1 + 2T          -1 - 4S\r\n" );
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document.write( "</PRE>\r<BR>\n" );
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document.write( "if the lines intersect, then the (x,y,z) coordinate of the intersection point will be the same in both lines.\r<BR>\n" );
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document.write( "this means that:\r<BR>\n" );
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document.write( "-1 + T = 4 + 3S<BR>\n" );
document.write( "6 - 2T = 2<BR>\n" );
document.write( "-1 + 2T = -1 - 4S\r<BR>\n" );
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document.write( "take any two of these equations and solve for one of the variables.<BR>\n" );
document.write( "then take the value of that variable and replace it in any one of the equations.\r<BR>\n" );
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document.write( "you will get S = -1 and T = 2\r<BR>\n" );
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document.write( "when S = -1 and T = 2, your lines will have a common x,y,z coordinate as shhown below.\r<BR>\n" );
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document.write( "for L1, x = -1 + 2 = 1<BR>\n" );
document.write( "for L2, x = 4 + 3 * -1 = 4 - 3 = 1\r<BR>\n" );
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document.write( "for L1, y = 6 - 4 = 2<BR>\n" );
document.write( "for L2, y = 2\r<BR>\n" );
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document.write( "for L1, z = -1 + 4 = 3<BR>\n" );
document.write( "for L2, z = -1 - 4 * -1 = -1 + 4 = 3\r<BR>\n" );
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document.write( "the common intersection occurs when T = 2 and S = -1.\r<BR>\n" );
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document.write( "that intersection point is at (1,2,3).\r<BR>\n" );
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document.write( "since i really didn't know how to do this when i decided to try to answer this question, i had to rely on many references to embellish my understanding sufficiently to provide an answer.\r<BR>\n" );
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document.write( "the other tutor who answered these questions was one of those references, namely math_helper(1240).\r<BR>\n" );
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document.write( "he was quite helpful.\r<BR>\n" );
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document.write( "the other references were on the web and are shown below.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"https://www.emathhelp.net/calculators/linear-algebra/angle-between-two-vectors-calculator/?ux=1&uy=-2&uz=2&vx=3&vy=0&vz=-4&steps=on\" target = \"_blank\">https://www.emathhelp.net/calculators/linear-algebra/angle-between-two-vectors-calculator/?ux=1&uy=-2&uz=2&vx=3&vy=0&vz=-4&steps=on</a>\r<BR>\n" );
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document.write( "there may be some redundancy, but better too much than too little so i included all that i found and reviewed that appeared to be pertinent to the problems at hand.\r<BR>\n" );
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document.write( "you may find them helpful as well.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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