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Unformatted text preview: WEEK TWO HW SOLUTIONS 1. Suppose you are the Captain of the USS Gauss, a brand new spacecraft, and you have just been launched into space to ﬂy to the moon. Your helper monkey Mojo has determined that the best landing spot on the moon has coordinates (1, −2, 3). You are currently at the point (−5, 3, 4) and ﬂying in the direction 2i − j. (a). How far are you from the landing spot? Solution. Let S (−5, 3, 4) be the current position of the spacecraft and M (1, −2, 3) be the landing spot on the moon. The distance between two points is given by: d(S, M ) = (1 − (−5))2 + (−2 − 3)2 + (3 − 4)2 = √ 36 + 25 + 1 = √ 62 (b). What is the parametric equation describing your ﬂight path? Solution. The parametric equations for a line starting at point S (−5, 3, 4) and with direction 2i − j are: x = −5 + 2t y =3−t for 0 ≤ t ≤ −∞ z=4 (c). Will you hit your landing spot? If not, how close will you get? Solution. The easiest way to see that you will not hit the landing spot is that the z coordinate of your trajectory is constant at 4 but the landing spot has a z coordinate of 3. There are multiple ways to ﬁgure out how close the ship will get to the landing spot. You can ﬁnd the distance between the point M (1, −2, 3) and the line from part (b), or you can minimize the distance between the point M (1, −2, 3) and the ship at any time t. Minimizing the distance d(S, M ) = (x − 1)2 + (y − (−2))2 + (z − 3)2 is equivalent to minimizing the distance squared [d(S, M )]2 = (x − 1)2 + (y − (−2))2 + (z − 3)2 Substitute the parametric equations for x, y , and z : [d(S, M )]2 = (−5 + 2t − 1)2 + (3 − t − (−2))2 + (4 − 3)2 = 5t2 − 34t + 62 Take the derivative and set equal to zero to ﬁnd the minimum: 10t − 34 = 0 ⇒ t = 17 5 This means the closest the ship will get is at t = 17 , where: 5 √ d(S, M ) = 5t2 − 34t + 62 = 5( 17 )2 − 34( 17 ) + 62 = 21 5 5 5
1 2 WEEK TWO HW SOLUTIONS (d). What path do you need to take to ensure that you hit the landing spot? Solution. To ensure that we hit the landing spot we need to travel in the direction of the vector −→ − SM = 6i − 5j − k Starting from S (−5, 3, 4) gives us the path r(t) = (−5 + 6t)i + (3 − 5t)j + (4 − t)k. Such that r(1) = i − 2j + 3k which is the position vector for M (1, −2, 3) 2. After successfully landing the USS Gauss on the moon, you begin your return trip home. You and Mojo are enjoying some freeze dried ice cream when suddenly, Mojo sees an astroid out the window. The astroid is located at the point (0, 5, −1) and has a heading of 2i + 1j + 3k. Your current ﬂight path is the line described by r(t) = (−1 + t)i + (2 − t)j. Will you hit the astroid? If not, how close will you get? (Hint: if two lines do not intersect, they can be embedded in parallel planes.) Solution. The path the asteroid is given by A(s) = (2s)i + (5 + s)j + (−1 + 3s)k and the path of the ship is r(t) = (−1 + t)i + (2 − t)j. Where the diﬀerent variables are to indicate that the asteroid and the ship might be traveling at diﬀerent speeds. Setting components of each path equal The zcoordinate: 0 = −1 + 3s ⇒ s = 1 3 The xcoordinate: 2s = −1 + t ⇒ t = 5 3 The ycoordinate: 5 + s = 2 − t ⇒ 16 = 1 3 3 Which means that these paths will not meet. Since the lines do not intersect they must lie in parallel planes. This means that the planes share a normal vector and to ﬁnd it we can use one vector from each plane in a cross product: Let v = 2i + 1j + 3k and the shuttle respectively. i jk 2 1 3 n=v×u= 1 −1 0 u = 1i − 1j be vectors corresponding to the paths of the asteroid and = 3i + 3j − 3k which we can scale to i + j − k → − To ﬁnd the distance between the lines is equivalent to ﬁnding the projection of any vector ( AS ) that runs from one line to the other onto the normal n. − → AS = (0 − ( 1))i (5 − 2)j + (3 − 0)k = i + 3j − k − + − → − → AS ·n projn AS = n·n n = ( 5 )(i + j − k) 3 The distance is now just the length of this projection: √ − → 5 projn AS  = ( 5 ) 3 = √3 3 3. Consider a room ten units long in the x, y , and z directions. Speciﬁcally, the walls of the room are the four planes, x = 0, x = 10, y = 0, and y = 10, and the ﬂoor and ceiling are z = 0 and z = 10, respectively. A ﬂat triangular mirror is mounted in one of the corners of the ceiling. The corners of the mirror are at locations (10, 9, 10), (10, 10, 9), and (9, 10, 10). You are sitting at location (5, 0, 0) playing with your new green laser pointer. WEEK TWO HW SOLUTIONS 3 (a). If you aim your laser pointer directly at the corner of the room with coordinates (10, 10, 10), determine the coordinates where the beam will hit the walls, or ﬂoor, of the room. (Hint: an incoming ray of light, and the surface normal where the ray hits the surface, form a plane. The reﬂected ray is in the same plane. Also, the angle between the incoming ray and the normal is the same as the angle between the normal and the reﬂected ray.) Solution. The hint that is given means that we can draw the following picture: Where LI is the incoming beam n is the normal to the mirror and LR is the reﬂected laser. These all lie in the same plane so ﬁnding LR is equivalent to reversing the perpendicular component of LI , i.e. L⊥ = −LI − projn (−LI ). First we need a normal vector to the mirror and a vector that points in the direction of the laser. If we use the three corners of the mirror P (10, 9, 10), Q(10, 10, 9), and S (9, 10, 10), we can obtain − − → − → two vectors in the plane of the mirror: P Q = j − k and P S = i − j, which we can cross to get the normal n: i j k − − →− → n = P Q × P S = 0 1 −1 = (0 − 1)i − (0 − (−1))j + (0 − 1)k = −i − j − k 1 −1 0 And for our incoming vector LI we want a vector from (5, 0, 0) to (10, 10, 10): LI = (10 − 5)i + (10 − 0)j + (10 − 0)k = 5i + 10j + 10k or LI = i + 2j + 2k for simplicity. Using the picture above you can see that LR = projn (−LI ) − L⊥ 4 WEEK TWO HW SOLUTIONS Where projn (−LI ) = And L⊥ = −LI − projn (−LI ) = −(i + 2j + 2k) − ( 5 )(−i − j − k) = 2 i − 1 j − 1 k 3 3 3 3 ⇒ LR = projn (−LI ) − L⊥ = ( 5 )(−i − j − k) − ( 2 i − 1 j − 1 k) = − 7 i − 4 j − 4 k 3 3 3 3 3 3 3 But since we only care about the direction we can scale this such that LR = −7i − 4j − 4k Now we have the direction of the reﬂected laser, but we still need to know where it starts, or the point that the incoming laser hits the mirror. Using the normal n = −i − j − k to the mirror, and a point on the mirror P (10, 9, 10) we have the standard form of an equation for a plane: −(x − 10) − (y − 9) − (z − 10) = 0 which can be rewritten as x + y + z = 29 We can also parametrize the line of the incoming laser that starts at point (5, 0, 0) with direction i +j + 2k by: 2 x=5+t y = 2t z = 2t It now remains to ﬁnd for what t the equation for the line satisﬁes the equation for the plane. Plugging in the parametric equation into the plane equation and solving for t: (5 + t) + (2t) + (2t) = 29 ⇒ t = 24 ⇒ x = 49 , y = 48 , z = 48 5 5 5 5 The laser hits the mirror at the point ( 49 , 48 , 48 ) and the reﬂected beam leaves with direction: 5 5 5 LR −7i − 4j − 4k Which give the following parametrization of the reﬂected beam: = x = 49 − 7t 5 y = 48 − 4t 5 z = 48 − 4t 5 All that is left is to ﬁgure out which “wall” the reﬂected laser will hit and where. Figuring out which wall is left to intuition and a little bit of guess work. If you choose the wrong wall you will ﬁnd that your answer does not make sense i.e. negative answers for the coordinates or other values “outside the box”. The correct “wall” ends up being the plane where x = 0, 0 ≤ y ≤ 10, and 0 ≤ z ≤ 10. This time we can use the fact that x = 0 in our parametrization of the reﬂected curve to ﬁnd t: 0 = 49 − 7t ⇒ t = 7 5 5 Using t = 7 gives us x = 0, y = 4, and z = 4. We have found that the reﬂected beam hits the 5 box at the point (0, 4, 4). (b). Suppose the ﬂat mirror is replaced with one octant of a spherical mirror of radius 1. The corners of the new spherical mirror are again located at (10, 9, 10), (10, 10, 9), and (9, 10, 10). If you again aim your laser pointer directly at the corner of the room with coordinates (10, 10, 10), determine the new coordinates where the beam will hit the walls, or ﬂoor, of the room. (Hint: a sphere has the property that at a point P on the surface, the normal to the surface, n, is parallel to the vector from the center of the sphere to the point P .) Solution. For this problem we need to see what angle the incoming laser makes with the normal at its point of impact. This is trivial though, since the laser is being directed at the center of the sphere, which means it is directed along the normal at the point that it hits. This diagram shows a cross section of the sphere and three diﬀerent incoming lasers and the resulting normals. −LI ·n n·n n = ( 5 )(−i − j − k) 3 WEEK TWO HW SOLUTIONS 5 Since the laser is aligned with the normal, the reﬂected beam will return right back to the starting point, (5, 0, 0). 4. Suppose you have the quadric surface Ax2 + By 2 + Cz 2 + F xz + K = 0. (a). What restrictions must you place on the coeﬃcients of your quadric surface to ensure the surface passes through the origin? Solution. For this surface to pass through the origin the point O(0, 0, 0) must satisfy the equation. Plugging in gives us: K = 0 which is our only restriction. (b). What restrictions must you place on the coeﬃcients of your quadric surface to ensure the surface passes through the yaxis? Solution. In order to pass through the yaxis there must be some point Y (0, y0 , 0) that satisﬁes the equation: 2 2 A(0)2 + By0 + C (0)2 + F (0)(0) + K = 0 ⇒ y0 = − K ⇒ y0 = ± − K B B which means we need
K B ≤ 0 for the surface to cross the yaxis (c). Assume that we label the octants of (x, y, z ) space in the following way: Octant I : x > 0, y > 0, z > 0 Octant II : x > 0, y < 0, z > 0 Octant III : x < 0, y < 0, z > 0 Octant IV : x < 0, y > 0, z > 0 Octant V : x > 0, y > 0, z < 0 Octant V I : x > 0, y < 0, z < 0 Octant V II : x < 0, y < 0, z < 0 Octant V III : x < 0, y > 0, z < 0 6 WEEK TWO HW SOLUTIONS Suppose you a have graph of your quadric surface in Octant I . In which other octants do you immediately know the shape of your surface? Now, suppose you have a graph of your surface in Octant III . In which other octants do you know the shape of your surface? Solution. Since the ﬁrst three terms are squared they will not change if you change the sign of x, y , or z . However the F xz term will depend on the signs of x and z . In Octant I , the value xz is positive which means we will know what the surface looks like where ever xz is also positive: In Octant II , Octant V II , and Octant V III . Similarly if we are given the data from Octant III we will know the graph when xz is negative: In Octant IV , Octant V , and Octant V I . (d). Suppose your surface passes through the origin. What conditions must be imposed on the coeﬃcients so that z = x, y = 0, is a line on the surface? If this line is on the surface, can the line z = −x, y = 0 also be on the surface? Explain why or why not. Solution. The surface passes through the origin so we can use the condition K = 0 from part (a). Plugging in the other conditions into the equation: Ax2 + By 2 + Cz 2 + F xz = 0 Ax2 + Cx2 + F xx = 0 ⇒ (A + C + F )x2 = 0 ⇒ we need A + C + F = 0 Similarly if we have z = −x, y = 0 on the surface we need to require that A + C − F = 0 In order to have both of the conditions we need that A + C = F and A + C = −F which can only happen if F = 0 and A + C = 0 which means that we can have both of these lines exist on the surface at once. (e). Now suppose your surface passes through the origin and intersects the line z = x, y = 0, and ﬁnally that A = C . You introduce the following change of variables x z x = √ − √ 2 2 x z z = √ + √ 2 2 Solve for x and z in terms of x and z . How are the x and z axes related to the x and z axes? (Hint: What do the lines z = x and z = −x become in the new coordinates?) Solution. Solving for x and z can be done by adding and subtracting the equations for x and z . 2 1 x + z = √2 x ⇒ x = √2 (x + z ) 2 1 z − x = √2 z ⇒ z = √2 (z − x )
1 1 The line x = z will become √2 (x + z ) = √2 (z − x ) ⇒ x = 0 1 1 And the line −x = z will become − √2 (x + z ) = √2 (z − x ) ⇒ z = 0 This means that the lines z = x and z = −x become our new axes in (x , y, z ) space. Another way to put this is that the transformation rotates space by 45◦ about the yaxis. WEEK TWO HW SOLUTIONS 7 (f ). Transform your expression for the quadric surface into a quadric surface in the variables (x , y, z ). Simplify the expression for your new quadric surface as much as possible. Solution. Starting from the general form Ax2 + By 2 + Cz 2 + F xz + K = 0 we place the restriction from the previous sections: K = 0, A = C , and A + C + F = 0 ⇒ 2A + F = 0 ⇒ F = −2A This gives us our reduced form: Ax2 + By 2 + Az 2 − 2Axz = 0 1 1 Substituting x = √2 (x + z ) and z = √2 (z − x ) 1 1 1 1 A( √ (x + z ))2 + By 2 + A( √ (z − x ))2 − 2A( √ (x + z ))( √ (z − x )) = 0 2 2 2 2 1 1 1 1 ⇒ A[ (x )2 + x z + (z )2 + (x )2 − x z + (z )2 − ((z )2 − (x )2 )] + By 2 = 0 2 2 2 2 2A y 2 2A or y = ± − x = B x B ⇒ 2A(x )2 + By 2 = 0 or − (g). Given your new expression for the quadric surface, what conditions must you enforce on your coeﬃcients such that you will have a surface at all? Assuming these conditions, draw your surface in the (x , y, z ) coordinate system. Then using your insight from above, draw your surface in the (x, y, z ) coordinate system. Solution. From the ﬁnal form above it is clear that we need to require that − 2A ≥ 0 in order to B have any surface at all. Setting A = −1 and B = 1 will yield the following graph in (x , y, z ) space: Simply rotating this 45◦ back will give us the following graph in (x, y, z ) space: 8 WEEK TWO HW SOLUTIONS ...
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This note was uploaded on 02/10/2011 for the course APPM 2350 taught by Professor Adamnorris during the Fall '07 term at Colorado.
 Fall '07
 ADAMNORRIS

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