Three-Dimensional Space

Three-Dimensional Space A line L in three-dimensional space is determined when we know a point P0 (x0 , y0 , z0 ) on L and the direction of L. In three dimensions the direction of a line is conveniently described by a vector, − − v be a vector parallel to L. Let P (x, y, z) be an arbitrary point on L and let → r0 and so we let → −−→ −−→ → − − r be the position vectors of P0 and P (that is, they have representations OP0 and OP ). If → a is −−→ → − → − − the vector with representation P0 P , then the Triangle Law for vector addition gives r = r0 + → a. → − → − → − → − But, since a and v are parallel vectors, there is a scalar t such that a = t v . Thus, → − − − r =→ r0 + t→ v, which is a vector equation of L. Exercise 1. Find a vector equation for the line that passes through (1, 0, −2) and is parallel to the vector <4,2,−1>. Definition: Two vectors are equal if and only if corresponding components are equal. Therefore, we have the three scalar equations: x = x0 + at y = y0 + bt z = z0 + ct where t ∈ R. These equations are called parametric equations of the line L through the point − P0 (x0 , y0 , z0 ) and parallel to the vector → v = < a, b, c >. Each value of the parameter t gives a point (x, y, z) on L. Exercise 2. Find parametric equations for the line L through P (5, −2, 4) that is parallel to → − 2 a < 1 2 , 2, −>. = 3 Exercise 3. Find parametric equations for the line through P1(3, 1, −2) and P2(−2, 7, −4). Exercise 4. Find parametric equations for the line through P (−3, 2, −3) and Q(1, −1, 4). Class Exercise 1. Find a vector equation and parametric equations for the line. (a) The line through the point (6,-5, 2) and parallel to the vector < 1, 3, − 32 > (b) The line through the point (0, 14,-10) and parallel to the line x = −1 + 2t, y = 6 − 3t, z = 3 + 9t Definition: T he s ymmetric e quations o f L are x−x0 a = y−y0 b = z−z0 c . Exercise 5. Find a symmetric form for the line through P1(3, 1, −2) and P2(−2, 7, −4). Class Exercise 2. Find parametric and symmetric equations for the line. (a) The line through the origin and the point (4, 3, −1). (b) The line through the points (1.0, 2.4, 4.6) and (2.6, 1.2, 0.3). → − → − → − → − (c) The line through (2,1,0) and perpendicular to both i + j and j + k − − Formula: The line segment from → r0 to → r1 is given by the vector equation → − − − r (t) = (1 − t)→ r0 + t→ r1 0 ≤ t ≤ 1. Exercise 6. Parametrize the line segment joining the points P (−3, 2, −3) and Q(1, −1, 4). Class Exercise 3. Find parametric equations for the line segment from (10, 3, 1) to (5, 6, −3). Definition: Skew lines are lines that do not intersect and are not parallel. Exercise 7. Show that the lines L1 and L2 with parametric equations: x=1+t y = −2 + 3t z=4−t x = 2s y=3+s z = −3 + 4s are skew. Class Exercise 4. Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If they intersect, find the points of intersection. (a) L1: x = 5 − 12t, y = 3 + 9t, z = 1 − 3t, L2: x = 3 + 8s, y = −6s, z = 7 + 2s y−3 z−2 x−2 z (b) L1 : x1 = y−1 −1 = 3 , L2 : 2 = −2 = 7 (#22) − Definition: A plane in space is determined by a point P0 (x0 , y0 , z0 ) in the plane and a vector → n → − that is orthogonal to the plane. This orthogonal vector n is called a normal vector. Definition: A vector equation of the plane is → − − − n ·(→ r −→ r0 ) = 0. Definition: A scalar equation of the plane through point P0 (x0 , y0 , z0 ) with normal vector → − n = < a, b, c > is a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. Exercise 8. Find an equation for the plane through P0 (−3, 0, 7) perpendicular to → − → − → − → − n = 5i + 2j − k. Exercise 9. Find a equation of the plane through the point (5, −2, 4) with normal vector = → − <1, a 2, 3>. Exercise 10. Find an equation for the plane through A(0, 0, 1), B(2, 0, 0), and C(0, 3, 0). Class Exercise 5. Find an equation of the plane. → − → − → − (a) The plane through the point (5,3,5) and with the normal vector 2 i + j − k (b) The plane through the point (2,0,1) and perpendicular to the line x = 3t, y = 2 − t, z = 3 + 4t (c) The plane through the point (2,4,6) and parallel to the plane z = x + y (d) The plane that contains the line x = 1 + t, y = 2 − t, z = 4 − 3t and is parallel to the plane 5x + 2y + z = 1 (e) The plane through the origin and the points (2,-4,6) and (5,1,3) (f) The plane that passes through the point (1,2,3) and contains the line x = 3t, y = 1 + t, z=2−t (g) The plane that passes through the point (1,-1,1) and contains the line with symmetric equations x = 2y = 3z Fact: If a, b, and c are not all 0, then the linear equation ax + by + cz + d = 0 represents a plane with normal vector < a, b, c >. Definition: T wo p lanes a re p arallel i f t heir n ormal vectors a re parallel. Exercise 11. Prove that the planes 2x − 3y − z − 5 = 0 and −6x + 9y + 3z + 2 = 0 are parallel. → − − Recall that if θ is the angle between two nonzero vectors → a and b , then cos θ = → − → − a·b → − . → − | a || b | Definition: I f two p lanes a re n ot p arallel, t hen t hey i ntersect i n a s traight l ine and the angle between the two planes is defined as the acute angle b etween their normal vectors. Exercise 12. Find the angle between the planes 3x − 6y − 2z = 15 and 2x + y − 2z = 5. Class Exercise 6. Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (a) 2z = 4y − x, 3x − 12y + 6z = 1 (b) 2x − 3y + 4z = 5, x + 6y + 4z = 3 (c) x + 2y + 2z = 1, 2x − y + 2z = 1 Distance Formula: The distance D from a point P1 (x1 , y1 , z1 ) to the plane ax + by + cz + d = 0: D= |ax1 +by1 +cz1 +d| √ . a2 +b2 +c2 Exercise 13. Find the distance from S(1, 1, 3) to the plane 3x + 2y + 6z = 6. Class Exercise 7. Find the distance between the point (−6, 3, 5) and the plane x − 2y − 4z = 8. Class Exercise 8. Find the distance between the given parallel planes: 6z = 4y − 2x and 9z = 1 − 3x + 6y.