Problem 3 Part a Because the line call it L1 is perpendicular to the plane call it P1 any normal vector to P1 can be a direction vector for L1 We know from the standard linear equation for a plane that one normal to P1 is 0 2 1 call it v1 Now we have a point on our line and its direction so we have a simple vector equation for L1 r1 t A tv1 2 3 1 t 0 2 1 Separating components we get our parametric equations x 2 t 0 2 y 3 t 2 3 2t z 1 t 1 1 t When looking for symmetric equations there is a pitfall the usual method of solving for t in all three variables is doomed because x has no dependence on t We can still solve for t as functions of y and z to get y 3 2 t 1 z The whole idea of the symmetric equations is to get rid of the dummy t variable so it is simpler to write this as y 3 2 1 z but we are not done This equation defines a plane we still have to add our restriction on x This is simple enough because we have already found our restriction on x Thus our symmetric equations are y 3 2 1 z x 2 1 Part b Because all lines are parallel to their direction vectors if we can show that the two lines direction vectors are parallel it follows that the lines themselves are parallel To find a direction vector of the second line call it L2 all we need to do is put together the components into a vector equation r2 t 0 2 1 t 0 4 2 Here we have a direction vector for L2 0 4 2 call it v2 Because v2 2v1 i e the direction vectors are scalar multiples of one another the two vectors are parallel and therefore the lines L1 and L2 are parallel to one another Another valid approach would be to show that v1 v2 0 so they must be parallel but this is more than you need Part c We know that a plane is defined by a point on it and a vector orthogonal to it i e a normal vector Because the plane we are looking for call it P contains every point in L1 and L2 it must contain A 2 3 1 so all we need is a normal vector We want to find two vectors w1 and w2 parallel to P but not parallel to one another If we then find a vector orthogonal to both w1 and w2 it will be orthogonal to every vector parallel to P and thus be a normal vector The simplest technique we know for finding a vector orthogonal to a pair of vectors is taking their cross product The central question then becomes how can we find two vectors parallel to the plane but not parallel to one another The natural temptation is to use the direction vectors of our two lines on P but from part b they are parallel so anything orthogonal to one is orthogonal to the other A simple way to find a vector parallel to a plane is to find the vector connecting two points on the plane However we have to 2 make sure that we don t pick two points that are either both on L1 or both on L2 because then the vector connecting them would be proportional and therefore parallel to the direction vectors of L1 and L2 We already know A is on L1 and we know that the two lines don t intersect after all they are parallel so we just need a point on L2 Plugging in t 0 in the parametric equations for L2 we find that 0 2 1 call it B lies on L2 Therefore we know that the vector 0 2 1 2 3 1 2 1 2 is parallel to P AB is not parallel to the direction vectors of the As predicted AB lines which are parallel to the plane Thus the cross product of and v1 v2 works just as well gives us a normal vector to the AB plane A quick calculation yields v1 5 2 4 AB Thus 5 2 4 is a normal vector of P All we need now is to plug our point A into the standard scalar equation for a plane to get our P This yields 5 x 2 2 y 3 4 z 1 0 This is a perfectly valid answer but it looks a little neater in linear form so we simplify our equation to 5x 2y 4z 0 It is always a good idea to test your final answer against points you know should work so we can plug in various points from our two lines and verify that this equation makes sense 5 2 2 3 4 1 10 6 4 0 5 0 2 2 4 1 0 4 4 0 5 0 2 0 4 0 0 0 0 0 5 2 2 5 4 0 10 10 0 0 This common sense check seems to confirm what we already showed somewhat more laboriously that the equation 5x 2y 4z 0 represents a plane containing both L1 and L2 3 Problem 6 Part a This is just a formula from the book b a b b 2 The idea behind the formula is that b a is simply the magnitude b of a times the cosine of a s angle with b When we multiply this b we get the length of the expression by the unit vector of b i e b b direction leg of the right triangle that has one leg in the direction of b and hypotenuse a Part b This is another formula from the book dy dy dx dx dt dt This intuitively makes sense if you think of the derivatives as fractions which isn t entirely accurate but it s a good guideline Then we have dy cos t dt dx sin t dt dy 4 cos 4 22 dt dx 4 sin 4 22 dt 2 dy 2 1 dx t 2 2 4 Notice that as it should be this is the slope of the path described by these parametric equations i e the unit circle at the given time 1 Part c FALSE One easy example is that the x and y axes are both perpendicular to the z axis but the x and y axes are not parallel to one another It is important to note that although this is sometimes false it is also sometimes true For instance if L is the x axis translated up by one in the z direction i e the line defined by the equations y 0 z 1 both L and the x axis are perpendicular to the z axis and L and the x axis are parallel to one another Part d TRUE If a line L is perpendicular to a plane P P s normal vectors are exactly L …