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1 Suggested problems - solutions Writing equations of lines and planes Some of these are similar to ones you have examples for... most of them aren t. P1: Write the general form of the equation of the plane containing the points (1, 3, 5), ( 2, 4, 1), and ( 1, 0, 1). Want: equation of a plane. Need: A point and a normal. Have: three points. Question becomes how to get normal. Relationship: Since the points are in the plane, we can write vectors v 1 and v 2 which lie in the plane as well. A normal to the plane is perpendicular to both vectors, so cross them. v 1 = AC =< 2, 3, 4 >. v 2 = AB =< 3, 1, 4 >. n = v 1 v 2 =< 16, 4, 11 >. Equation of plane given by n (r r 0 ) = 0, where r =< x, y, z >, and r 0 is position vector of any point on the plane; I ll use A for r 0 =< 1, 3, 5 >. < 16, 4, 11 > < x 1, y 3, z 5 > = 0 16(x 1) + 4(y 3) 11(z 5) = 0 16x y 12 11z + 55 = 0 16x + 4y 11z + 27 = 0 Final equation is in general form. You can check by verifying all three points satisfy the equation.

2 P2: Write the symmetric equations of a line through the point (1, 4, 1) and parallel to the line with symmetric equations x 2 = y = z 2 Want: equation of line parallel to given line. Need: a point and a direction vector for that line. Have: the point, and the given parallel line. Question becomes how to get direction vector for wanted line. Relationship: Direction vector for given L1 read off the equation. Since parallel, the lines are in the same direction: v 2 = v 2 =< 3, 4, 2 >. Point-parallel formula is < x, y, z >=< x 0, y 0, z 0 > +t < a, b, c >, so < x, y, z >=< 1, 4, 1 > +t < 3, 4, 2 > Since the problem requests final answer in symmetric form, solve it out: x = 1 + 3t t = x 1 3 y = 4 4t t = y 4 4 z = 1 2t t = z x 1 3 = y 4 4 = z + 1 2

3 P3: Write the point-normal form of the equation of a plane containing the line and the vector < 1, 1, 1 >. L : < x, y, z >=< 4, 1, 0 > +t < 2, 3, 4 > Want: equation of plane. Need: a point in the plane and a normal. Given: a line and a vector both in the plane. Relationship: The line is geometrically in the plane in space - any point on the line is a point on the plane. (You can t say this about vectors in a plane, by the way, because they represent the difference between points and move around in space.) So, any arbitary point on the line works as our point for the plane. Say (4, 1, 0). The direction vector for the line, v 1 =< 2, 3, 4 >, is in the plane. The other given vector v 2 =< 1, 1, 1 > is also in the plane. So the normal to the plane must be orthogonal to both; cross them. Point-normal equation: Stop there - that s the requested form. n = v 1 v 2 =< 7, 2, 5 > < 7, 2, 5 > < x 4, y 1, z 0 >= 0

4 P4: Write the point-parallel equation of a line through the point (1, 2, 1) and orthogonal to a plane containing the vectors < 1, 3, 2 > and < 10, 0, 1 >. Want: equation of a line. Need: point on line and direction vector for line. Have: vectors in a plane orthogonal to the line. Question becomes how to get direction vector for line - how is it related to normal for plane? Relationship: Since the line is orthogonal to the plane, it is parallel to the plane s normal. Direction vector v for the line is the same as normal n, and n can be found by crossing the vectors in the plane: Equation (point-parallel): v = n = v 1 v 2 =< 3, 21, 30 > < x, y, z >=< 1, 2, 1 > +t < 3, 21, 30 >

5 P5: Write the general form of the equation of the plane through the point (0, 0, 5) and parallel to the plane x y + 3z 6 = 0. Want: equation of plane. Need: point and normal. Have: point given. Have another parallel plane. How are normals related? Relationship: Parallel planes have parallel normals. n 2 = n 1 =< 1, 1, 3 >. Point-normal, solve to general: < 1, 1, 3 > < x 0, y 0, z 5 > = 0 1(x 0) 1(y 0) + 3(z 5) = 0 x y + 3z 15 = 0

6 P6: Write the general form of the equation of a plane through the point (0, 0, 5) and perpendicular to the plane x y + 3z 6 = 0. How many planes meet this description? Want: equation of plane. Need: point and normal. Have: point, and another perpendicular plane. How are normals related? Relationship: Since planes are orthogonal, so are the normals: n 1 n 2. That s not enough to fix a single direction; there are an infinite number of possibilities here. Since we need a vector orthogonal to only one given vector, we need a solution to n 1 n 2 = 0. Letting n 2 =< a, b, c >: < 1, 1, 3 > < a, b, c >= 0 a b + 3c = 0 Say a = 1, b = 2, so c = 1. The vector n 2 =< 1, 2, 1 > is orthogonal to the vector n 1 =< 1, 1, 3 >. Equation (solve to general): < 1, 2, 1 > < x 0, y 0, z 5 > = 0 x + 2y + z 5 = 0

7 P7: Find parametric equations for the line through the point (0, 1, 2) that is parallel to the plane x + y + z = 2 and orthogonal to the line x = 1 + t, y = 1 t, z = 2t. Want: equation of line. Need: point and direction vector. Have: point and parallel plane and orthogonal line. How is direction vector for line related to these two things? Relationship: v 2 v 1 (since lines are orthogonal). v 2 n (since lines parallel to plane, and thus orthogonal to normal). v 2 orthogonal to two known vectors, so cross them: Equation (solve to parametric): v 2 = n v 1 =< 1, 1, 1 > < 1, 1, 2 >=< 3, 1, 2 > < x, y, z >=< 0, 1, 2 > +t < 3, 1, 2 > x = 3t y = 1 t z = 2 2t

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