SOLUTIONS TO HOMEWORK ASSIGNMENT #2, Math 253

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1 SOLUTIONS TO HOMEWORK ASSIGNMENT #, Math 5. Find the equation of a sphere if one of its diameters has end points (, 0, 5) and (5, 4, 7). The length of the diameter is (5 ) + ( 4 0) + (7 5) = =, so the radius is. The centre is at the midpoint ( +5, 0 4, 5+7 ) = (,, ). Hence, the sphere is given as (x ) + (y + ) + (z ) = 9.. Find vector, parametric, and symmetric equations of the following lines. (a) the line passing through the points (,, ) and (4,, ) The vector between two points is v = 4,, =, 4, 5. Hence the equation of the line is Vector form: r = r 0 + t v = 4,, + t, 4, 5 = 4 + t, 4t, + 5 t Parametric form: x = 4 + t, y = 4t, z = + 5 t Symmetric from: Solving the parametric form for t gives x 4 = y+ 4 = z 5/ (b) the line passing through the origin and perpendicular to the plane x 4y = 9 Perpendicular to the plane parallel to the normal vector n =, 4, 0. Hence Vector form: r = 0, 0, 0 + t, 4, 0 = t, 4t, 0 Parametric from : x = t, y = 4t, z = 0 Symmetric form x = y 4, z = 0 (c) the line lying on the planes x + y z = and x 4y + 5z = We can find the intersection (the line) of the two planes by solving z in terms of x, and in terms of y. () x + y z = () x 4y + 5z = Solve z in terms of y: () () 7y 8z = 0 z = 7 8 y Solve z in terms of x: 4 () + () 7x + z = 4 z = 4 7x Hence, symmetric form: 4 7x = 7 8 y = z Set the symmetric form = t, we have parametric form: x = 4 t 7, y = 8 7 t, z = t Vector form: r = 4 t, 8 t, t 7 7

2 . Find the equation of the following planes. (a) the plane passing through the points (,, ), (,, ), and (4, 0, ) Name the points P (,, ), Q(,, ), and R(4, 0, ). Set up two vectors: v = P Q = +,, + =,, () v = P R = 5,, 4 () Choose the normal vector n = v v = 5, 7, 8. Hence the equation of the plane is 5(x + ) + 7(y ) + 4(z + ) = 0 using point P. (b) the plane passing through the point (0,, ) and containing the line x = y = z Name Q(0,, ). The line can be represented as r = t, t, t, which crosses the point P (0, 0, 0) and is parallel to v =,,. Set b = P Q = 0,,. Choose n = v b =,, and hence the equation of the plane is x y + z = 0 using point P. (c) the plane containing the lines L : x = + t, y = t, z = 4t L : x = s, y = + s, z = 4 + s From L and L, v =,, 4 and v =,,. Choose n = v v = 9, 5,. Since L crosses the point (,,0), the equation of the plane is 9(x ) 5(y ) + z = Find the intersection of the line x = t, y = t, z = t, and the plane x + y + z =. Substitute the line into the plane: t + t + t = t =. Put t back to the line: x =, y =, z =. Hence the intersection point is (,, ). 5. Find the distance between the point (, 8, 5) and the plane x y z =. Name Q(, 8, 5). Choose any point on the plane, say a convenient one (x, 0, 0). So x (0) (0) = x = P (, 0, 0). Then b = P Q =, 8, 5. The normal vector of the plane is n =,,. The distance between the plane and the point is given as

3 distance = proj n b n b = n = 5 = 5. Show that the lines are skew. L : x 4 L : x Write the equation in parametric form. = y = y + = z = z L : x = t + 4, y = 4t 5, z = t + L : x = s +, y = s, z = s The lines are not parallel since the vectors v =, 4, and v =,, are not parallel. Next we try to find intersection point by equating x, y, and z. () t + 4 = s + () 4t 5 = s () t + = s () gives s = t +. Substituting into () gives 4t 5 = (t + ) t = 5. Then s = 8. However, this contradicts with (). So there is no solution for s and t. Since the two lines are neither parallel nor intersecting, they are skew lines. 7. Identify and sketch the following surfaces. (a) 4x + 9y + z = xy-plane: 4x + 9y = ellipse xz-plane: 4x + z = ellipse yz-plane: 9y + z = ellipse ellipsoid (b) 4z x y = xy-plane: x y = nothing, try z = constants z = c: x y = 4c x + y = 4c circles when 4c > 0 xz-plane: 4z x = hyperbola opening in z-direction yz-plane: 4z y = hyperbola opening in z-direction hyperboloid of two sheets

4 (c) y = x + z xy-plane: y = x cross xz-plane: 0 = x + z point at origin, try y = constants y = c: c = x + z circles yz-plane: y = z cross cone (d) x + 4z y = 0 xy-plane: x y = 0 y = x parabola opening in +y-direction xz-plane: x + 4z = 0 point at origin, try y = constants y = c: x + 4z c = 0 x + 4z = c ellipses when c > 0 yz-plane: 4z y = 0 y = 4z parabola opening in +y-direction elliptic paraboloid (e) y + 9z = 9 x missing: cylinder along x-direction yz-plane: y + 9z = 9 ellipse elliptic cylinder (f) y = z x xy-plane: y = z parabola opening in +y-direction xz-plane: 0 = z x z = x cross, try y = constants y = c: c = z x hyperbola opening in z-direction when c > 0, in x-direction when c < 0 yz-plane: y = x parabola opening in y-direction hyperbolic paraboloid 8. Find the polar equation for the curve represented by the following Cartesian equation. (a) x = 4 x = 4 r cos θ = 4 r = 4 sec θ (b) x + y = x x + y = x r = r cos θ r = cos θ 4

5 z 0 0 y 0x Figure : Q7(a) 0 y 0 x Figure : Q7(b) 0 y 0 x Figure : Q7(c) 0 y 0 x Figure 4: Q7(d) 0 0 x y Figure 5: Q7(e) z y 0x Figure : Q7(f) (c) x y = x y = r cos θ r sin θ = r (cos θ sin θ) = r cos θ = r = sec θ r = ± sec θ 9. Sketch the curve of the following polar equations. (a) r = 5 (b) θ = π 4 (c) r = sin θ (d) r = ( cos θ) 0. (a) Change (, π, ) from cylindrical to rectangular coordinates x = r cos θ = cos π =, y = r sin θ = sin π =, z =. Hence (x, y, z) = (,, ) (b) Change (,, 4) from rectangular to cylindrical coordinates r = x + y = + =, tan θ = y = x θ = π in first quadrant, z = 4. Hence (r, θ, z) = (, π, 4) 5

6 Figure 7: Q9(a) Figure 8: Q9(b) Figure 9: Q9(c) Figure 0: Q9(d) (c) Change (,, ) from rectangular to spherical coordinates ρ = x + y + z = + + = 4, tan θ = y = x θ = π in first quadrant, φ = cos z = ρ cos = cos = π. Hence (ρ, θ, φ) = (4, π, π) 4 (d) Change (4, π, π ) from spherical to cylindrical coordinates 4 r = ρ sin φ = 4 sin π =, θ = π, z = ρ cos φ = 4 cos π 4 (, π 4, ) =. Hence (r, θ, z) =