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Topic 4: Vectors: Summary Information LINES IN 3 DIMENSIONS l Consider a line in 3 dimensions passing through the point (a, b, c) parallel to the vector m n Cartesian equation of a line: Vector equation of a line xa y b z c a l l m n r b t m c n or r = (ai + bj + ck) + t(li + mj + nk) Example: Find the equation of the line passing through the points A(3, 5, 2) and B(2, -4, 5). Solution: 2 3 1 Find the direction of the line: One possible direction vector is AB 4 5 9 5 2 3 x3 y 5 z 2 The Cartesian equation of this line is (using the coordinates of point A). 1 9 3 x 3 1 The equivalent vector equation is y 5 t 9 . z 2 3 PLANES IN 3 DIMENSIONS There are various alternative ways to write the equation of a plane: The equation of a plane passing through the point Vector equation of a plane ( a1 , a2 , a3 ) , in the direction of the two vectors a1 b1 c1 b1 c1 r a2 t b2 s c2 b and c is … a b c 2 2 3 3 3 b c3 or r = a + tb + sc 3 The equation of a plane perpendicular to the vector Cartesian equation of a plane e n f , is … ex + fy + gz = d g or r .n d (n is called the normal vector) The equation of a plane perpendicular to the vector e (r a). n 0 n f and passing through the point (a1 , a2 , a3 ) is g or r.n a.n THE SCALAR PRODUCT (or DOT PRODUCT) The scalar product is a way of combining two vectors together to get a scalar quantity. Key formula: a . b a b cos where a is the magnitude of vector a b is the magnitude of vector b θ is the angle between vectors a and b. a1 b1 The scalar product is found using a . b a b a b a b 2 2 1 1 2 2 3 3 a b 3 3 Properties of the scalar product: 1) a.b=b.a 2) If a and b are perpendicular then a . b = 0 (and vice versa) THE VECTOR PRODUCT (or CROSS PRODUCT) The vector product is a way of combining two vectors together to get another vector quantity. Key formula: a b (a b sin )n ˆ ˆ where n is a vector perpendicular to both a and b a is the magnitude of vector a b is the magnitude of vector b θ is the angle between vectors a and b. a1 b1 i a1 b1 The vector product is found using a b det j a b2 2 2 k a2 a b 3 3 3 b3 Properties of the scalar product: 1) a × b = - b × a 2) a × b gives a vector perpendicular to both of the original vectors. Example: 1 2 Find a vector perpendicular to both 4 and 0 . 2 3 Solution: We find the vector product of these two vectors: 1 2 i 1 2 12 4 0 det j 4 0 i 4 0 j 1 2 k 1 2 7 . 2 3 k 2 3 2 3 2 3 4 0 8 We can check our answer by finding the scalar product of the answer with each of the original vectors. We should get 0. COMMON QUESTIONS FINDING THE CARTESIAN EQUATION OF A PLANE GIVEN 3 POINTS Example: Find the Cartesian equation of the plane passing through the points (2, 0, -1), (3, 1, 2) and (1, 2, 5). Solution: There are 2 methods we can use: Method 1: Finding the vector equation and then eliminating the parameters Two possible direction vectors are: 3 2 1 1 2 1 1 0 1 and 2 0 2 2 1 3 5 1 6 Therefore a possible vector equation is: x 2 1 1 y 0 1 2 z 1 3 6 To find this equation in Cartesian form, we eliminate the parameters from the equations. We first form 3 equations: x 2 i.e. x2 (1) y 0 2 i.e. y 2 (2) z 1 3 6 i.e. z 1 3 6 (3) We can eliminate λ by finding (2) – (1): y – x + 2 = 3μ 1 1 1 2 i.e. ( y x 2) y x 3 3 3 3 1 1 2 1 2 4 From equation (2), we have: y 2 y 2 y x y x 3 3 3 3 3 3 1 2 4 1 1 2 Substituting into equation (3) gives: z 1 3 y x 6 y x 3 y 3 3 3 3 3 3 This gives the equation of the planes as 3y – z = 1 Method 2: Finding a normal vector using the vector product We start by finding the vector product of the two direction vectors: 1 1 i 1 1 0 1 2 = det j 1 2 i 1 2 j 1 1 k 1 1 0i 9 j 3k 9 3 6 k 3 6 3 6 3 6 1 2 3 This normal vector gives the coefficients of x, y and z in the equation of the plane. Therefore we know that the plane has equation 0x – 9y + 3z = d. Substitute in the coordinates of one of the points, e.g. (1, 2, 5), to get -18 + 15 = d i.e. d = -3 So the planes has equation -9y + 3z = -3 OR 3y – z = 1 FINDING THE LINE OF INTERSECTION OF TWO PLANES Two non-parallel planes intersect to form a line. We can find the equation of that line as follows: Example: Find the line of intersection of the planes 2x –y + 2z = -6 x + 2y + z = 2 Solution: Because there are 3 unknowns but only two equations, we introduce a parameter t. The equations then are 2x – y + 2z = -6 x + 2y + z = 2 z=t Substituting the 3rd equation into the top two equations and rearranging gives: 2x – y = -6 – 2t (1) x + 2y = 2 – t (2) Multiply the top equation by 2: 4x – 2y = -12 – 4t x + 2y = 2 – t Add to eliminate y: 5x = -10 – 5t i.e. x = -2 – t. Substitute this into equation (1): 2(-2 – t) – y = -6 – 2t -4 –2t = -6 – 2t + y y = 2. The solutions therefore are x = -2 - t, y = 2 and z = t. Writing this in vector form, we get: x 2 t 2 1 y 2 2 t 0 z t 0 1 2 1 i.e. r 2 t 0 0 1 1 This is the equation of a line through the point (-2, 2, 0), in the direction of the vector 0 . 1 FINDING THE INTERSECTION OF A PLANE AND A LINE There are several possibilities: the line could lie within the plane; the line could intersect the plane at a single point; the line could be parallel to the plane. Example: Investigate which of the three situations above applies with the line x 5 y 9 z 1 1 6 2 and the plane 2x – y + 4z = 5. Solution: To investigate this situation, it is simplest if we have the equation of the line in vector form and the equation of the plane in Cartesian form (which it is already). x 5 1 The vector equation of the line is: y 9 t 6 z 1 2 i.e. x=5–t y = 9 + 6t z = 1 + 2t Substituting these into the equation of the plane gives: 2x – y + 4z = 5 2(5 – t) – (9 + 6t) + 4(1 + 2t) = 5 10 – 2t – 9 – 6t + 4 + 8t = 5 5=5 This equation is true for all values of t. Therefore the line must lie within the plane. Note 1: If when we attempt to solve the equations, we get an equation which is true for no values of t (such as 4 = -2), we know that the plane and the line do not intersect, i.e. are parallel. Note 2: If when we attempt to solve the equations, we get an equation which is true for a single value of t, we know that the plane and the line intersect at a single point. We can find the point of intersection by substituting the value of t into the vector equation of the line. FINDING THE ANGLE BETWEEN TWO PLANES The angle between two planes is the same as the angle between the two normal vectors. 3 1 Example: Find the acute angle between the planes with equations: r. 0 2 and r. 2 1 1 5 Solution: The equations of the planes are 3x + 0y – z = 2 and x + 2y + 5z = -1. 3 1 The normal vectors are 0 and 2 . 1 5 3 1 The scalar product is: 0 . 2 (3 1) (0 2) (1 5) 2 1 5 3 1 The magnitudes are: 0 9 0 1 10 and 2 1 4 25 30 1 5 Substituting into the formula a.b=abcosθ gives: 2 2 10 30 cos cos 96.6 10 30 The acute angle between these planes therefore is: 180 – 96.6 = 83.4°. FINDING THE ANGLE BETWEEN A PLANE AND A LINE To find the angle between a plane and a line, there are 2 steps: Step 1: Find the angle between the normal vector and the direction vector of the line; Step 2: Subtract the angle from step 1 from 90° in order to get the angle required. 4 x2 y z 3 Example: Find the angle between the line and the plane r. 3 5 . 2 1 1 2 Solution: 2 4 First we find the angle between the vectors 1 and 3 : 1 2 2 4 1 . 3 8 3 2 13 1 2 2 4 1 4 11 6 3 16 9 4 29 1 2 Therefore: 13 6 29 cos 9.76 So the angle between the plane and the line is 90 – 9.76 = 80.2°. FINDING THE PERPENDICULAR DISTANCE FROM A POINT TO A LINE Example: Find the perpendicular distance from the point P(3, 5, 2) to the line with equation 2 1 r 3 t 1 . 1 4 Solution: The line passes through the point A(2, 3, -1) and has direction i – j + 4k. Let Q be the closest point on the line to P. P(3, 5, 2) Then PQ is perpendicular to the direction of the line. 1 –1 As Q lies on the line, the position vector of Q will take the form: 4 2 1 2 t Q OQ 3 t 1 3 t 1 4 1 4t A(2, 3, -1) for some t. 2 t 3 1 t Therefore PQ q p 3 t 5 2 t . 1 4t 2 3 4t Since PQ is perpendicular to the direction of the line, we have: 1 t 1 2 t . 1 0 3 4t 4 1 t 2 t 12 16t 0 18t 11 11 t 18 So, by substituting this value for t into the expression for PQ , we get 1 18 11 7 11 1 PQ 2 18 47 3 44 10 18 18 The shortest distance from point P to the line is the magnitude of this vector: 7 1 1 PQ 47 49 2209 100 2.70 18 18 10 FINDING THE PERPENDICULAR DISTANCE FROM A POINT TO A PLANE Example: Find the perpendicular distance from the point P(3, 5, 2) to the plane with equation 3x – 2y + z = 4 P(3, 5, 2) Solution: Let Q be the foot of the perpendicular from point P to the plane. Then PQ will be perpendicular to the plane. 3 However, the normal vector 2 is a vector 1 Q (x, y, z) perpendicular to the plane. So 3x - 2y + z = 4 3 PQ t 2 . 1 Therefore the equation of the line PQ is 3 3 r 5 t 2 . 2 1 We can find the coordinates of Q by seeing where this line intersects the plane 3x – 2y + z = 4. Substituting x = 3 + 3t, y = 5 – 2t, z = 2 + t into the equation of the plane, gives: 3(3 + 3t) – 2(5 – 2t) + (2 + t) = 4 9 + 9t – 10 + 4t + 2 + t = 4 1 + 14t = 4 3 So t . 14 3 So the vector PQ is PQ 3 2 . 14 1 Therefore the perpendicular distance from P to the plane is 3 PQ 14 2 14 9 4 1 0.802. 3 3 1 Note: There is a formula (quoted in the formula book) which can also be used to find the perpendicular distance from a point to a line. See the next page. FINDING THE PERPENDICULAR DISTANCE FROM A POINT TO A PLANE (FORMULA BOOK METHOD) Formula quoted in the OCR formula book: The perpendicular distance of ( , , ) from 1 x 2 y 3 z d 0 is 1 2 3 d 12 2 32 2 Example (continued): Find the perpendicular distance from the point P(3, 5, 2) to the plane with equation 3x – 2y + z = 4 Solution: ( , , ) = (3, 5, 2). The equation of the plane is 3x – 2y + z – 4 = 0. Therefore the perpendicular distance is: 1 2 3 d (3 3) ( 2 5) (1 2) 4 3 = 0.802 12 2 32 2 32 (2)2 12 14 FINDING THE SHORTEST DISTANCE BETWEEN TWO SKEW LINES You need to learn the following formula for finding the shortest distance between two skew lines: If two lines l1 and l2 are given by r a1 tb1 and r a 2 sb 2 respectively, the shortest distance d between l1 and l2 is given by: (a2 a1 ).n d where n b1 b 2 n 2 1 3 2 Example: Find the shortest distance between the lines r 3 t 1 and r 1 s 2 5 2 1 3 Solution: 1 2 i 1 2 1 n 1 2 j 1 2 i 1 2 j 1 2 k 1 2 7 2 3 k 2 3 2 3 2 3 1 2 4 3 2 1 1 1 (a 2 a1 ).n 1 3 . 7 4 . 7 1 28 24 53 1 5 4 6 4 Therefore the shortest distance is 53 53 6.52 1 49 16 66

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posted: | 8/26/2011 |

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