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Practice Finding Planes and Lines in R Here are several main types of problems you ﬁnd in 12.5 and old exams pertaining to ﬁnding lines and planes: LINES 1. Find an equation for the line that goes through the two poin ts (1 2) and (4 3). 2. Find an equation for the line that is parallel to the line = 3 = 6 = 7 + 2 and goes through the point (0 2). 3. Find an equation for the line that is orthogonal to the plan e 3 + 2 = 10 and goes through the point (1 2). 4. Find an equation for the line of intersection of the plane 5 = 4 and 10 = 6. PLANES 1. Find the equation of the plane that goes through the three p oints (0 4), (1 0), and 4). 2. Find the equation of the plane that is orthogonal to the lin = 4 + = 1 = 8 and goes through the point (3 1). 3. Find the equation of the plane that is parallel to the plane +2 = 6 and goes through the point (4 2). 4. Find the equation of the plane that contains the intersect ing lines = 4 + = 2 = 1 and = 4 = 3 = 1 + 2 5. Find the equation of the plane that is orthogonal to the pla ne 3 + 2 = 4 and goes through the points (1 4) and 2). LINES/PLANES/SPHERES AND INTERSECTIONS 1. Find the intersection of the line = 3 = 1 + 2 = 2 and the plane 2 + 3 = 4. 2. Find the intersection of the two lines = 1 + 2 = 3 = 5 and = 6 = 2 + 4 = 3 + 7 (or explain why they don’t intersect). 3. Find the intersection of the line = 2 = 3 and the sphere = 16. 4. Find the intersection of the plane 3 = 0 and the sphere = 4.

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LINES (Solutions) 1. (a) A position vector: (b) A direction vector: 2) (c) Equation: which gives = 1 + 3 = 0 2 + 5 2. (a) A position vector: (b) A direction vector: (Parallel to the other line, so we can use the same direction vector). (c) Equation: which gives = 0 = 1 + 6 = 2 + 7 3. (a) A position vector: (b) A direction vector: (Orthogonal to the plane, so we can use the normal from the plane). (c) Equation: which gives = 1 + 3 = 4 2 + 2 4. Solution Method 1 : Find two points of intersection. There are many point we jus t need to ﬁnd two. (a) First let’s combine and simplify. Adding the equations g ives 15 + 2 = 10 (b) Pick some numbers. If = 0, then we get 2 = 10, so = 5. And going back to the original equations and plugging in (to either one) we get 0 + 5 + = 4, so 1. Hence, (0 1) is a point on the line we desire. If = 0, then we get 15 = 10, so = 2 3. And going back to the original equation we get 5(2 3) + 0 + = 4, so = 4 10 3 = 2 3. Thus another point is (2 3). You can check that these points work in both equations. Now we can use the standard line method. (c) A position vector: (d) A direction vector: (e) Equation: which gives = 0 + 2 = 5 Solution Method 2 : Find one point of intersection then use the cross-produce o f the normal for the direction. (a) For this method you still have to ﬁnd one point of intersec tion. So for example (0 1) as we did above. (b) The cross product of the normals for each plane will give a vector that is parallel to the line (picture it). So this is another way to get a direction vector . That would give 10 10 10) 10 20 (c) A position vector: (d) A direction vector: 20 . (Note this is parallel to the direction vector we got with method 1). (e) Equation: which gives = 0 = 5 + 20 . Remember you and your classmate may have diﬀerent parameterizations and bot h be correct. But your direction vectors should be parallel.

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PLANES (Solutions) 1. (a) A position vector: (b) A normal vector: AB and AC , so one normal vector is 12 (c) Equation: ) = 0 which gives 12( 0) + 4( 3) + 2( 4) = 0, or more simply 12 + 4 + 2 20 = 0. 2. (a) A position vector: (b) A normal vector: (Orthogonal to the line, so the direction vector for the line is a normal to the plane). (c) Equation: ) = 0 which gives ( 3) 2( 2) + 8( 1) = 0, or more simply + 8 7 = 0. 3. (a) A position vector: (b) A normal vector: (Parallel to the other plane, so same normal works). (c) Equation: ) = 0 which gives 5( 4) 3( + 1) + 2( 2) = 0, or more simply + 2 27 = 0. 4. Note that the lines intersect at = 0 and = 0, which gives the point (4 1). We can quickly ﬁnd three points by also plugging in = 1 and = 1 which gives (5 2) and (1 3). So we have three points. Note also that PQ and PR (so I really didn’t have to ﬁnd and I could have just grabbed the direction vectors from the line s). (a) A position vector: (b) A normal vector: 13 (c) Equation: ) = 0 which gives 13( 4) + 7( + 0) + 9( 1) = 0, or more simply 13 + 7 + 9 61 = 0. 5. You have two vectors parallel to the plane. One is PQ and the other is the normal from the given plane which is (a) A position vector: (b) A normal vector: (c) Equation: ) = 0 which gives 3( 1) 8( 2) 7( 4) = 0, or more simply + 41 = 0.

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LINES/PLANES/SPHERES AND INTERSECTIONS (Solutions) 1. (a) Combine and ﬁnd 2(3 ) + 3(1 + 2 (2 ) = 4 gives 6 + 3 + 6 2 + = 4, so 13 = 3 and = 3 13. (b) Get the point: Thus, = 9 13, = 1 + 6 13 = 29 13, and = 2 13 = 23 13. 2. (a) Combine and ﬁnd and i. 1 + 2 = 6 implies that = 5 ii. 3 = 2 +4 combined with the fact just obtained gives 3 = 2 + 4(5 ) which gives = 22 , so 11 = 22 Hence, = 2 and going back, we also get = 1. Thus, the only parameters that simultaneously work to equate and are = 2 and = 1. Now we check the third equation. iii. 5 = 3 + 7 . Plugging in = 2 and = 1 we get 10 = 3 + 7, it works! (b) Get the point: Thus, = 5, = 6, and = 10 is the point where the two lines intersect. 3. (a) Combine and ﬁnd (2 + (3 + ( = 16 gives 4 + 9 + 4 = 16, so 17 = 16 and 16 17 = 17. (b) Get the points: Thus, the two points of intersection are (8 17 12 17 17) and 17 12 17 17). 4. (a) Combine Since we get + ( = 4 which gives + 10 = 4. (b) What is this: So every point that satisﬁes +10 = 4 with is a point of intersection. That is really the best we can do. (In terms of looking from abo ve, meaning the projection onto the xy -plane, +10 = 4 would look like an ellipse. Also, is a plane through the origin and if you visualize the intersection you will see that it is just a great circle of the sphere). In any case, the point is that the intersection of tw o surfaces is typically a curve in two dimensions, not just a point.

5 and old exams pertaining to 64257nding lines and planes LINES 1 Find an equation for the line that goes through the two poin ts 1 2 and 4 3 2 Find an equation for the line that is parallel to the line 3 6 7 2 and goes through the point 0 2 3 Fi ID: 23049

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Practice Finding Planes and Lines in R Here are several main types of problems you ﬁnd in 12.5 and old exams pertaining to ﬁnding lines and planes: LINES 1. Find an equation for the line that goes through the two poin ts (1 2) and (4 3). 2. Find an equation for the line that is parallel to the line = 3 = 6 = 7 + 2 and goes through the point (0 2). 3. Find an equation for the line that is orthogonal to the plan e 3 + 2 = 10 and goes through the point (1 2). 4. Find an equation for the line of intersection of the plane 5 = 4 and 10 = 6. PLANES 1. Find the equation of the plane that goes through the three p oints (0 4), (1 0), and 4). 2. Find the equation of the plane that is orthogonal to the lin = 4 + = 1 = 8 and goes through the point (3 1). 3. Find the equation of the plane that is parallel to the plane +2 = 6 and goes through the point (4 2). 4. Find the equation of the plane that contains the intersect ing lines = 4 + = 2 = 1 and = 4 = 3 = 1 + 2 5. Find the equation of the plane that is orthogonal to the pla ne 3 + 2 = 4 and goes through the points (1 4) and 2). LINES/PLANES/SPHERES AND INTERSECTIONS 1. Find the intersection of the line = 3 = 1 + 2 = 2 and the plane 2 + 3 = 4. 2. Find the intersection of the two lines = 1 + 2 = 3 = 5 and = 6 = 2 + 4 = 3 + 7 (or explain why they don’t intersect). 3. Find the intersection of the line = 2 = 3 and the sphere = 16. 4. Find the intersection of the plane 3 = 0 and the sphere = 4.

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LINES (Solutions) 1. (a) A position vector: (b) A direction vector: 2) (c) Equation: which gives = 1 + 3 = 0 2 + 5 2. (a) A position vector: (b) A direction vector: (Parallel to the other line, so we can use the same direction vector). (c) Equation: which gives = 0 = 1 + 6 = 2 + 7 3. (a) A position vector: (b) A direction vector: (Orthogonal to the plane, so we can use the normal from the plane). (c) Equation: which gives = 1 + 3 = 4 2 + 2 4. Solution Method 1 : Find two points of intersection. There are many point we jus t need to ﬁnd two. (a) First let’s combine and simplify. Adding the equations g ives 15 + 2 = 10 (b) Pick some numbers. If = 0, then we get 2 = 10, so = 5. And going back to the original equations and plugging in (to either one) we get 0 + 5 + = 4, so 1. Hence, (0 1) is a point on the line we desire. If = 0, then we get 15 = 10, so = 2 3. And going back to the original equation we get 5(2 3) + 0 + = 4, so = 4 10 3 = 2 3. Thus another point is (2 3). You can check that these points work in both equations. Now we can use the standard line method. (c) A position vector: (d) A direction vector: (e) Equation: which gives = 0 + 2 = 5 Solution Method 2 : Find one point of intersection then use the cross-produce o f the normal for the direction. (a) For this method you still have to ﬁnd one point of intersec tion. So for example (0 1) as we did above. (b) The cross product of the normals for each plane will give a vector that is parallel to the line (picture it). So this is another way to get a direction vector . That would give 10 10 10) 10 20 (c) A position vector: (d) A direction vector: 20 . (Note this is parallel to the direction vector we got with method 1). (e) Equation: which gives = 0 = 5 + 20 . Remember you and your classmate may have diﬀerent parameterizations and bot h be correct. But your direction vectors should be parallel.

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PLANES (Solutions) 1. (a) A position vector: (b) A normal vector: AB and AC , so one normal vector is 12 (c) Equation: ) = 0 which gives 12( 0) + 4( 3) + 2( 4) = 0, or more simply 12 + 4 + 2 20 = 0. 2. (a) A position vector: (b) A normal vector: (Orthogonal to the line, so the direction vector for the line is a normal to the plane). (c) Equation: ) = 0 which gives ( 3) 2( 2) + 8( 1) = 0, or more simply + 8 7 = 0. 3. (a) A position vector: (b) A normal vector: (Parallel to the other plane, so same normal works). (c) Equation: ) = 0 which gives 5( 4) 3( + 1) + 2( 2) = 0, or more simply + 2 27 = 0. 4. Note that the lines intersect at = 0 and = 0, which gives the point (4 1). We can quickly ﬁnd three points by also plugging in = 1 and = 1 which gives (5 2) and (1 3). So we have three points. Note also that PQ and PR (so I really didn’t have to ﬁnd and I could have just grabbed the direction vectors from the line s). (a) A position vector: (b) A normal vector: 13 (c) Equation: ) = 0 which gives 13( 4) + 7( + 0) + 9( 1) = 0, or more simply 13 + 7 + 9 61 = 0. 5. You have two vectors parallel to the plane. One is PQ and the other is the normal from the given plane which is (a) A position vector: (b) A normal vector: (c) Equation: ) = 0 which gives 3( 1) 8( 2) 7( 4) = 0, or more simply + 41 = 0.

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LINES/PLANES/SPHERES AND INTERSECTIONS (Solutions) 1. (a) Combine and ﬁnd 2(3 ) + 3(1 + 2 (2 ) = 4 gives 6 + 3 + 6 2 + = 4, so 13 = 3 and = 3 13. (b) Get the point: Thus, = 9 13, = 1 + 6 13 = 29 13, and = 2 13 = 23 13. 2. (a) Combine and ﬁnd and i. 1 + 2 = 6 implies that = 5 ii. 3 = 2 +4 combined with the fact just obtained gives 3 = 2 + 4(5 ) which gives = 22 , so 11 = 22 Hence, = 2 and going back, we also get = 1. Thus, the only parameters that simultaneously work to equate and are = 2 and = 1. Now we check the third equation. iii. 5 = 3 + 7 . Plugging in = 2 and = 1 we get 10 = 3 + 7, it works! (b) Get the point: Thus, = 5, = 6, and = 10 is the point where the two lines intersect. 3. (a) Combine and ﬁnd (2 + (3 + ( = 16 gives 4 + 9 + 4 = 16, so 17 = 16 and 16 17 = 17. (b) Get the points: Thus, the two points of intersection are (8 17 12 17 17) and 17 12 17 17). 4. (a) Combine Since we get + ( = 4 which gives + 10 = 4. (b) What is this: So every point that satisﬁes +10 = 4 with is a point of intersection. That is really the best we can do. (In terms of looking from abo ve, meaning the projection onto the xy -plane, +10 = 4 would look like an ellipse. Also, is a plane through the origin and if you visualize the intersection you will see that it is just a great circle of the sphere). In any case, the point is that the intersection of tw o surfaces is typically a curve in two dimensions, not just a point.

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