ha lvh262 Homework 12 5 karakurt 56295 This print out should have 17 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 Which of the following surfaces is the graph of 4x 3y 6z 12 in the first octant z 10 0 points Which of the following statements are true for all lines and planes in 3 space 1 1 I two lines parallel to a third line are parallel II two planes perpendicular to a third plane are parallel y III two lines perpendicular to a plane are parallel x z 1 I and III only correct 2 2 II only 3 I only 4 all of them y 5 II and III only x 6 III only z 7 I and II only 3 8 none of them Explanation I TRUE each of the two lines has a direction vector parallel to the direction vector of the third line so must be scalar multiples of each other II FALSE the xy plane and yz plane are both perpendicular to the xz pane but are perpendicular to each other not parallel y x z 4 III TRUE the two lines will have direction vectors parallel to the normal vector of the plane and so be parallel hence the two lines are parallel 002 y 10 0 points x ha lvh262 Homework 12 5 karakurt 56295 2 z z 5 y x y x z 6 coras its graph in the first octant 1 y x 2 rect 3 Explanation Since the equation is linear it s graph will be a plane To determine which plane we have only to compute the intercepts of 4 5 4x 3y 6z 12 6 Now the x intercept occurs at y z 0 i e at x 3 similarly the y intercept is at y 4 while the z intercept is at z 2 By inspection therefore the graph is z y x 3 x 5 x 3 x 4 x 4 x 5 y 4 y 3 y 5 y 3 y 5 y 4 z 5 z 4 z 4 z 5 z 3 z 3 1 correct 1 1 1 1 1 Explanation As the surface is a plane it must be the graph of a linear function which can be written in intercept form as x y z 1 a b c But by inspection we see that the x intercept is x 3 the y intercept is y 4 and the zintercept is z 5 Consequently the surface is the graph in the first octant of the equation x y z 1 3 4 5 x 003 10 0 points Which equation has the surface 004 10 0 points Find parametric equations for the line passing through the point P 4 4 1 and parallel to the vector h 1 1 4 i ha lvh262 Homework 12 5 karakurt 56295 1 x 4 t correct y 4 t z 1 4t 2 x 1 4t y 1 4t z 4 t 3 x 4 t y 4 t z 1 4t 4 x 1 4t y 1 4t z 4 t 5 x 1 4t y 1 4t z 4 t 6 x 4 t y 4 t z 1 4t Explanation A line passing through a point P a b c and having direction vector v is given parametrically by r t a tv a h a b c i Now for the given line a h 4 4 1 i v h 1 1 4 i Thus r t h 4 t 4 t 1 4t i 3 4 Q 0 7 2 5 Q 0 1 6 6 Q 7 6 0 Explanation Since the xy plane is given by z 0 we have to find an equation for and then set z 0 Now a line passing through a point P a b c and having direction vector v is given parametrically by r t a tv a h a b c i But for a h 3 2 1 i v h 4 4 1 i Thus r t h 3 4t 2 4t 1 t i so z 0 when t 1 Consequently the line intersects the xy plane at Q 1 2 0 Consequently x 4 t y 4 t z 1 4t keywords line parametric equations direction vector point on line intercept coordinate plane are parametric equations for the line 006 keywords line parametric equations direction vector point on line 005 10 0 points A line passes through the point P 3 2 1 and is parallel to the vector h 4 4 1 i At what point Q does intersect the xyplane 1 Q 2 1 0 2 Q 1 2 0 correct 3 Q 0 6 7 10 0 points Find parametric equations for the line passing through the point P 1 3 1 and perpendicular to the plane 2x y 3z 2 1 x 1 2t y 3 t z 1 3t 2 x 2 t y 1 3t z 3 t 3 x 1 2t y 3 t z 1 3t 4 x 1 2t y 3 t z 1 3t correct ha lvh262 Homework 12 5 karakurt 56295 5 x 2 t y 1 3t z 3 t 6 x 2 t y 1 3t z 3 t Explanation A line passing through a point P a b c and having direction vector v is given parametrically by r t a tv a h a b c i Now for the given line its direction vector will be parallel to the normal to the plane 2x y 3z 2 Thus Explanation Since the xy plane is given by z 0 we have to find an equation for and then set z 0 Now a line passing through a point P a b c and having direction vector v is given parametrically by r t a tv v h 2 1 3 i and so a h a b c i But for its direction vector will be parallel to the normal to the plane 4x 2y 4z 2 Thus a h 1 1 4 i a h 1 3 1 i 4 v h 4 2 4 i and so r t h 1 4t 1 2t 4 4t i r t h 1 2t 3 t 1 3t i Consequently In this case z 0 when t 1 Consequently intersects the xy plane at Q 3 1 0 x 1 2t y 3 t z 1 3t are parametric equations for the line 007 10 0 points A line passes through the point P 1 1 4 and is perpendicular to the plane 4x 2y 4z 2 At what point Q does intersect the xyplane 1 Q 3 0 5 2 Q 5 3 0 3 Q 1 3 0 4 Q 5 0 1 5 Q 3 0 3 6 Q 3 1 0 correct keywords line parametric equations direction vector point on line intercept coordinate plane 008 10 0 points Find parametric equations for the line passing through the points P 4 2 …