MATH 00 (Fall 016) Exam 1 Solutions 1 1. (a) (10 points) Find an equation of the sphere with center (, 1, 4). (x ( )) + (y 1) + (z ( 4)) 3 (x + ) + (y 1) + (z + 4) 9 (b) (10 points) Find an equation of the sphere thru the point (, 1, 4) which passes thru the origin. Corrected: Find an equation of the sphere with center (, 1, 4) that passes through the origin. The radius of this sphere must be ( 0) + (1 0) + ( 4 0) + 1 + 4 4 + 1 + 16 1. (x + ) + (y 1) + (z + 4) 1
MATH 00 (Fall 016) Exam 1 Solutions. (a) (15 points) Find an equation of the plane that passes through the point (3,, 1) and is perpendicular to the line x 6t, y + 4t, z 3t. The line is parallel to the vector 6, 4, 3. Since the plane is perpendicular to the line, any vector normal to the plane must be parallel to the line. In particular, one normal vector to the plane is n 6, 4, 3. Then an equation for the plane may have any of the following forms: n x 3, y ( ), z 1 0, 6, 4, 3 x 3, y +, z 1 0, 6(x 3) + 4(y + ) + 3(z 1) 0, 6x + 4y + 3z + ( 18 + 8 3) 0, 6x + 4y + 3z 13 0, 6x + 4y + 3z 13. (b) (5 points) The plane that you computed in part (a) intersects the xyplane in a line. What is the equation of that line? Solution 1 Let (x, y, z) be a point on this line. Then (x, y, z) is on the plane from part (a), so 6x + 4y + 3z 13. But this point is also on the xy-plane, so z 0. Hence in two dimensions the line has the equation which may be rewritten as 6x + 4y 13, y 6 4 x + 13 4 3 x + 13 4.
MATH 00 (Fall 016) Exam 1 Solutions 3 Solution Let (x, y, z) be a point on this line. Then (x, y, z) is on the plane from part (a), so 6x + 4y + 3z 13. Hence in two dimensions the line has the equation which may be rewritten as 6x + 4y 13, y 6 4 x + 13 4 3 x + 13 4. A vector representation in three dimensions for this line is r(t) t, 3 t + 13 4, 0 Parametric equations are t, 3 t, 0 + 0, 13 4, 0 t 1, 3, 0 + 0, 13 4, 0. x t y 3 t + 13 4 z 0.
MATH 00 (Fall 016) Exam 1 Solutions 4 Solution 3 The line must have the equation r(t) b + tv, where b and v are three-dimensional vectors to be determined. Let n 1 6, 4, 3 be a vector normal to the plane from part (a) and n k 0, 0, 1, which is normal to the xy-plane. Then v must be parallel to n 1 n, so we may as well set v n 1 n i j k 6 4 3 0 0 1 4 3 0 1 i 6 3 0 1 j + 6 4 0 0 k ( 4(1) 3(0) ) i ( 6(1) 3(0) ) j + ( 6(0) 4(0) ) k (4 0)i (6 0)j + 0k 4i 6j 4, 6, 0. A point (x, y, z) on the line must satisfy z 0 (since the point is in the xy-plane) and 13 6x + 4y + 3z 6x + 4y since the point is on the plane from part (a). There are (infinitely) many possibilities for such a point; one option is to set x 0, in which case Thus a point on the line is y 13 4. ( 0, 13 ) 4, 0,
MATH 00 (Fall 016) Exam 1 Solutions 5 so we may take b 0, 13 4, 0. An equation for the line is then r(t) 0, 13 4, 0 + t 4, 6, 0. Parametric equations are x 4t y 13 4 6t z 0. Compare these parametric equations with the ones given in Solution 1.
MATH 00 (Fall 016) Exam 1 Solutions 6 3. (0 points) Determine whether the following two lines are parallel, skew, or intersect. If they intersect, find the point of intersection. L 1 : x 7 6t 1, y 4, z 5 + t 1 L : x t, 1 + t, z + 5t The line L 1 is parallel to the vector 6, 0,. The line L is parallel to the vector 1, 1, 5. The vectors 6, 0, and 1, 1, 5 are not parallel. If they were parallel, there would be a scalar α such that 6, 0, α 1, 1, 5. Equating the second components, this would force 0 α(1) α. But then 6, 0, 0 ( 1, 1, 5) 0, 0, 0, which is impossible. Now we check if L 1 and L intersect. We must solve the system for two numbers t 1 and t. 7 6t 1 t 4 1 + t 5 + t 1 + 5t There are many possible ways to solve this system. One approach is to use the second equation to find t 3 and then obtain from the first equation 7 6t 1 3 5 6t 1 1 t 1. Now we check if the lines do intersect, which amounts to checking 5 + t 1? + 5t when t 1 and t 3.
MATH 00 (Fall 016) Exam 1 Solutions 7 But whereas 5 + t 1 5 + () 5 + 4 9, + 5t + 5(3) + 15 13, so the z-coordinates do not agree, and the lines do not intersect. Another way to phrase this is that L 1 has vector equation and L has vector equation We have and Since r 1 (t) 7, 4, 5 + t 6, 0,, r (t), 1, + t 1, 1, 5. r 1 () 7, 4, 5 + 6, 0, 7, 4, 5 + 1, 0, 4 5, 4, 9 r (3), 1, + 3 1, 1, 5, 1, + 3, 3, 15 5, 4, 13. 5, 4, 9 11, 4, 13, we have r 1 () r (3), so the lines cannot intersect. Since the lines are not parallel and do not intersect, they are skew.
MATH 00 (Fall 016) Exam 1 Solutions 8 4. (a) (1 points) Find the equation of the tangent line to the graph of r(t) e t i + tj (sin(t))k at the point where t. You may write your answer either in the form of a vector equation or as parametric equations. Let r(t) e t, t, sin(t). r() e,, sin() e,, 0 r (t) r () e t, 1, cos(t) t e 1,, cos() e, 1, 1 The vector equation for the tangent line is l(t) r() + tr () e,, 0 e,, 0 + t Parametric equations are e 1,, 1. x e + te y + t z t.
MATH 00 (Fall 016) Exam 1 Solutions 9 (b) (8 points) Given that v, 1, 3 and b 1,,, find the vector component of v along b and the vector component of v orthogonal to b. Clearly indicate which one is along b and which one is orthogonal to b. The vector component of v along b is ( ) ( ) v b, 1, 3 1,, proj b (v) b b 1,, 1,,. We compute this as ( ) (1) + ( 1)() + 3() proj b (v) 1,, 1 + + + 6 1,, 1 + 4 + 4 6 1,, 9 1,, 3 3, 4 3, 4. 3 The vector component of b orthogonal to b is v proj b (v), 1, 3 3, 4 3, 4 3 6 3 3, 3 3 4 3, 9 3 4 3 4 3, 7 3, 5. 3
MATH 00 (Fall 016) Exam 1 Solutions 10 5. (a) (10 points) Find the solution to the following initial value problem: dr dt t, 0, sin(t), r(0) 1,, 3. We obtain three scalar-valued initial value problems: { r 1(t) t, { r (t) 0, { r 3(t) sin(t), r 1 (0) 1 r (0) r 3 (0) 3. We solve each scalar IVP by direct integration: r 1 (t) t + C 1, r (t) C, r 3 (t) cos(t) + C 3. We use the initial conditions to find the constants of integration: Hence r 1 (0) 1 0 + C 1 1 C 1 1, r (0) 0 + C, r 3 (0) 3 cos(0) + C 3 3 1 + C 3 3 C 3 4. t r(t) + 1,, cos(t) + 4.
MATH 00 (Fall 016) Exam 1 Solutions 11 (b) (10 points) The curve r(t) cos(t), t, sin(t) is a helix. For the value t /4, compute the equation of the tangent line to the curve. Where does the tangent line cross the xz-plane? r ( ( 4) cos ) 4, 4, sin ( ) 4 1, 4, 1 r (t) sin(t), 1, cos(t) r ( ( 4) sin ( 4), 1, cos ) 4 1, 1, 1 The equation of the tangent line at t /4 is ( ) ( ) l(t) r + tr 4 4 1, 4, 1 + t 4 + t 0 t 4. Hence the tangent line cross the xz-plane at the point ( l ) 4 1, 1, 1, 4, 1 ( + ) 1, 1, 4 1 1, 4, 1 + 4, 4, 4 4 + 4, 0, 4 4. 1
MATH 00 (Fall 016) Exam 1 Solutions 1 6. (5 point bonus) Find the plane through the point ( 1, 4, ) which contains the line of intersection for the planes 8x y + z 4 0 and 4x + y 4z 6 0. You can check that the point P 0 ( 1, 4, ) does not lie on either of the planes, so let us find two points P 1 and P that lie on both planes. Then a vector normal to our desired plane will be n P 0 P 1 P 1 P. To find these points P 1 and P, we need to solve the pair of equations 8x y + z 4 0 4x + y 4z 6 0. If we add these equations together, they become 1x z 10 0 6x z 5 0 z 6x 5. If x 0, then z 5, and the equation for the first plane becomes So P 1 (0, 7, 5). y 10 4 0 y 14 0 y 7. If x 1, then z 1 and the equation for the first plane becomes 8(1) y +(1) 4 0 8 y + 4 0 6 y 0 y 3. So P (1, 3, 1). Then and P 0 P 1 0 ( 1), 7 4, 5 1, 11, 7 P 0 P 1 ( 1), 3 4, 1, 1, 1. Then (omitting the work for the cross product) P 0 P 1 P 0 P 4, 13, 1. 1 point: Hence an equation for the plane is 4, 13, 1 x ( 1), y 4, z 0, which may be rewritten in many ways, including 4(x + 1) 13(y 4) + 1(z ) 0.