Math 234 Practice Test 1 Show your work in all the problems 1 Find parametric equations for the line in which the planes x 2y z 1 and x y 2z 8 intersect 2 Compute the distance from the point 2 2 3 to the plane through the points A 0 0 0 B 2 0 1 and C 2 1 0 3 Compute the area of the parallelogram with three of its vertices given by A 2 2 1 B 3 1 2 and C 3 1 1 4 Cancellation in a dot product Let u v w be three vectors with u 6 0 Is it true that u v u w implies v w If you think it is true explain why otherwise provide a counterexample 5 Sketch the surface given by the equation z 1 x2 6 Describe the given sets with a single equation or a pair of equations The circle of radius 1 centered at 3 4 1 and lying in a plane parallel to the a xy plane b yz plane c xz plane 1 Solutions 1 Normal vectors of the two planes are given by n1 1 2 1 and n2 1 1 2 respectively The line of intersection is perpendicular to both n1 and n2 The following vector is then parallel to the line of intersection n1 n2 i j k 1 2 1 1 1 2 1 2 1 1 2 1 k j i 1 1 1 2 1 2 5i j 3k 5 1 3 We also need to find a point which lies on the line of intersection We insert x 1 2y z which is derived from the first equation into the second equation and we get 8 x y 2z 1 2y z y 2z 1 3y z as well as z 9 3y and x 1 2y z 10 5y This means we can pick any value we want for y and compute x z using the previous formulas For y 0 we get x 10 and z 9 so that the point 10 0 9 lies on the line of intersection Parametric equations of the line are then given by x 10 5t y t z 9 3t Comment There are many possible solutions which are all correct and for which you would get full credit For example the vector 5 1 3 is also parallel to the line of intersection do you know why Later on instead of choosing y 0 to find a point on the line we could have chosen something else for example y 1 We would have obtained x 10 5y 5 and z 9 3y 6 so that we use the point 5 1 6 instead Remember that there are many points on a line 2 and there is no reason to prefer one over the other The parametric equations would then look as follows x 5 5t y 1 t z 6 3t Although these equations are different they describe the same line and they are therefore also a valid solution to the problem 2 We need to find a normal vector n to the plane i e a vector perpen dicular to both AB 2 0 1 and AC 2 1 0 We get a such a vector by taking the cross product i j k n AB AC 2 0 1 i 2j 2k 2 1 0 If P is any point in the plane then the distance d between S 2 2 3 and the plane is given by PS n d n The easiest pick for P is probably 0 0 0 so that P S 2 2 3 We get P S n 2 2 3 1 2 2 2 4 6 12 and so that n q 1 2 2 2 2 2 9 3 12 4 3 3 The area of the parallelogram is given by d AB AC We have i j k AB AC 1 1 1 i j 1 1 0 1 1 0 and AB AC 3 2 4 The statement is false Pick two different vectors v and w for example v 1 0 0 and w 0 1 0 Then choose u perpendicular to both v and w for example u 0 0 1 would do Then u v u w 0 but v 6 w 5 The surface is a cylinder over the parabola z 1 x2 in the xz plane 6 a x 3 2 y 4 2 1 z 1 b y 4 2 z 1 2 1 x 3 c x 3 2 z 1 1 1 y 4 4