Vectors, Lines And Planes - THE LOVE WEDDING SHOOT
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesVectors, Lines and PlanesPrerequisites: Adding, subtracting and scalar multiplying vectors;calculating angles between vectors.Maths Applications: Describing geometric transformations.Real-World Applications: Description of forces; solar sailing; rotationalmechanics; planetary orbits.Direction Ratios and Direction CosinesThis short section contains some simple ideas that are useful in latersections.Definition:The direction ratio of a vector is the ratio of its components in orderfrom first to last. p Thus, the direction ratio of p 1 is p1 : p2 and the direction ratio of p2 q1 q q2 is q1 : q2 : q3 . q 3 Theorem:Vectors have equal direction ratios if and only if they are parallel.Example 1 2 2 Determine which of the vectors defined by p 1 , q 1 and r 3 3 4 2 are parallel. 6 M Patel (April 2012)1St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesWorking out the direction ratios of each vector (and simplifying), we get,p1 : p2 : p3 2 : 1 : 3q1 : q2 : q3 2 : 1 : 3r1 : r2 : r3 2 : 1 : 3Hence, as the direction ratios of all 3 vectors are equal, all 3 vectors areparallel to each other.Definition:If α , β and γ are the angles the vector p makes with the x, y and z axesrespectively, and u is a unit vector in the direction of p, the directioncosines of p are cos α , cos β and cos γ and satisfy, cos α u cos β cos γ cos 2 α cos 2 β cos 2 γ 1In other words, the direction cosines of a vector p are the components ofa unit vector in the direction of p.Example 2Find the direction cosines of the vector p 6 i 18 j - 22 k.The magnitude of p is p 62 182 ( 22)2 844 . Hence, a unitvector in the direction of p is, 6 844 18 u 844 22 844 M Patel (April 2012)2St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planesand thus the direction cosines of p are cos α cos γ 228446844, cos β 18844and.The Vector ProductVector FormDefinition:The vector product (aka cross product) of 2 vectors a and b, where θis the angle from a to b, and n is a unit vector at right angles to both aand b, is defined as,defa x b a b sin θ nImagine grasping a pole with your right hand. Then the direction in whichyour fingers curl around the pole indicate the direction of the angle froma to b and the thumb points in the direction of n.Note that while the scalar product can be defined in any number ofdimensions, the vector product is a 3D-only quantity. Also, the order inwhich the vector product is taken matters (as opposed to the scalarproduct) – figure out why.Also note that the vector product is a vector and is at right angles toboth a and b.The vector product arises in many areas of physics, especially rotationalmechanics.Example 3Calculate the vector product afollowing diagram.M Patel (April 2012)b, as well as its magnitude, in thex3St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes3b60 2aUsing the definition of the vector product,a x b a b sin θ nwhere n points upwards, gives,a x b 3 2 sin 60 n 3 32nExample 4Calculate the vector product afollowing diagram.b, as well as its magnitude, in thex3a60 2bWe have,a x b 3 32nwhere n points upwards.M Patel (April 2012)4St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesGeometrical Interpretation of the Vector Product MagnitudeThe magnitude of the vector product has a geometric interpretation interms of area. By considering the following diagram,bθawe see that the following theorem is true.Theorem:The area of a parallelogram with side lengths a and b is a x b .Theorem:The 3 unit vectors, i, j and k satisfy,i x j kj x k ik x i jandi x iM Patel (April 2012) j x j 5k x k 0St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesThe Scalar Triple ProductDefinition:The scalar triple product of 3 vectors a, b and c is,[a, b, c] a (b x c)def a1 a2 a3 b1 b2 b3 c c c 1 2 3 Example 5Work out the scalar triple product [b, a, c] for the vectorsa 3 i 2 j - k, b - 2 i 5 k and c - i j 4 k.We have,[b, a, c] b (a x c) b1 b2 b3 a1 a2 a3 c c c 1 2 3 2 0 5 3 2 1 1 1 4 18 0 25 7The scalar triple product has a geometrical interpretation as the volumeof a parallelepiped (roughly speaking, a 3D parallelogram). Make a sketch !M Patel (April 2012)6St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesTheorem:The volume of a parallelepiped with side lengths a , b and c is given byany of the following 6 expressions,a (b x c) a x (b c)b (c x a) b x ( c a)c (a x b) c x (a b)This theorem is useful in proving the last 2 properties in the nextsubsection.Properties of the Vector Product a x a 0 a x b -b x a a x (b c) (a b) x c (a x c) (b x c) (a x b) (a x c)Example 6 2 For a 3 , a x a a a sin 0 n 0. 1 Notice that the ‘zero’ is the zero vector 0 and not the zero scalar 0.Example 7Show that (a - b) x (a b) 2 (a x b).(a - b) x (a b) (a x a) (a x b) - (b x a) - (b x b) 0 (a x b) (a x b) - 0M Patel (April 2012)7St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes 2 (a x b)Theorem:If the vector product of 2 vectors is 0, then they are parallel.Example 8Show that any vector is parallel to itself.As a x a 0, a is parallel to itself.Theorem:If 2 non-zero vectors are parallel, then their vector product is 0.Example 9Show that a x (2a b) (b x a) 0.a x (2a b) (b x a) 2a x a (a x b) (b x a) 0 (a x b) - (a x b) 0Component FormTheorem:Given 2 vectors a a1 a2 and b a 3 b1 b2 , the component form of the b 3 vector product is,a x b ( a2b3 a3b2 ) i ( a3b1 a1b3 ) j ( a1b2 a2b1 ) kM Patel (April 2012)8St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesExample 10Calculate the vector product of a 2 i - 5 j 3 k and b 4 i - 11j - 7 k.a x b ( 35 33 ) i ( 12 14 ) j ( 22 20 ) k 68 i 26 j - 2 kExample 11Find a unit vector in the direction of s x t, where s and t - 9 i 3 k.i 2j 3 ks x t 6 i - 30 j 18 kThe magnitude of s x t is 6 35 . Hence, a unit vector in the direction ofs x t is 1 35 5 335 35 Example 12Calculate the exact value of the sine of the angle between the vectorss i j 2 k and t - 5 j k.s x t s t sin θ nWe have s x t 11 i - j - 5 k, s 6 and t 26 . A unit vector 11 7 3 in the direction of s x t is n 7 1 3 . Hence, 5 7 3 M Patel (April 2012)9St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes 11 1 5 11 7 3 26 sin θ 7 1 3 5 7 3 6Taking the scalar product of each side of the above equation with n gives,149 626 sin θ7 3 626 sin θ7 3sin θ 72 13Example 13Show that no value of p makes the vectors a 2p i j - 3 k andb - 7 i p j - k parallel.For the vectors to be parallel, we require a x b 0. We have, 3p 1 a x b 2p 21 2p 2 7 The third component obviously cannot equal 0. Hence, this vector clearlycannot equal 0. Thus the given vectors cannot possibly be parallel.Equations of PlanesThe equation of a plane can be found in 3 ways, depending on theinformation given. 1 point in the plane and a vector at right angles to the plane. 2 vectors in the plane. 3 points in the plane.M Patel (April 2012)10St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesDefinition:A vector is parallel to a plane if it lies in the plane.Definition:A normal vector to a plane is one that is at right angles to any vector inthe plane.Definition:The Cartesian equation of a plane containing a point P (x, y, z) is,ax by cz d(a, b, c, d, x, y, z ℝ)The equation of a plane is not unique. Multiplying the equation of a planeby a non-zero scalar will give an apparently different equation, but it’sreally the same (as the scalar can be divided out).Example 14Determine whether or not the planes 2x 5y6x 15y 9z 12 are the same. 3z 4 andAs the second plane equation is just 3 times the first plane equation, the2 planes are identical.Example 15Determine whether the points P (2, 0, 0) and Q (1, 1, 1) lie on the plane2x 5y 3z 4.As 2(2) 5(0)As 2(1) 5(1)M Patel (April 2012) 3(0) 4, P lies on the plane. 3(1) 0 4, Q doesn’t lie on the given plane.11St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesTheorem:A normal vector to the plane ax by cz d is, a n b c Note that a plane has 2 possible directions for a normal. The other a normal for the above plane would be b . c Theorem:Parallel planes have the same direction ratios for their normals.Cartesian Equation of a Plane Given Normal and 1 Point on PlaneGiven a normal and 1 point on the plane, the Cartesian equation of theplane can be quickly found by determining d.Example 16Determine the Cartesian equation of the plane passing through the point 2 (2, 3, 1) and which has normal vector 0 . 1 Substituting the given point and normal components into the genericCartesian equation gives d 2(2) 3(0) 1( 1) 3. Hence, theCartesian equation is,2x z 3M Patel (April 2012)12St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesCartesian Equation of a Plane Given 2 Vectors and a Point in thePlaneBy taking the vector product of 2 vectors lying in the plane, a normal canbe determined. Together with the given point that lies on the plane, weare back to the previous case.Example 17Find the equation of the plane passing through (3, 1, 1) and containing the 2 2 vectors 5 and 1 . 1 1 A normal to the plane is, 6 2 2 5 x 1 4 8 1 1 Then d 3( 6) 1(4) equation is, upon simplification,3x 2y1( 8) 22. Hence, the Cartesian 4z 11Cartesian Equation of a Plane Given 3 Points in the PlaneGiven 3 points on a plane, 2 vectors emanating from one point can beformed and thus the equation can be determined from the previous case.Example 18Determine the equation of the plane containing the points A (1, 6, 0),B ( 4, 2, 5) and C ( 2, 4, 1). The vectors AB and AC are,M Patel (April 2012)13St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes 5 3 AB 8 and AC 10 5 1 Taking the vector product gives us a normal to the plane, 58 AB x AC 20 26 Then d 58(1) 20( 6) 26(0) 62. Hence, an equation for theplane is,58x 20y 26z 62which becomes, upon simplification,29x 10y 13z 31Notice that the vector product can be taken in any order; the negativesign will cancel out from the plane equation.Definition:For a point A (with position vector a) in a plane, and 2 vectors b and cparallel to the plane, a vector equation (aka parametric equation) of aplane for a point R (with position vector r), where t and u are realparameters, is,r a t b u cNote that the vectors b and c are parallel to (in) the plane. If we are told3 points that lie in the plane, then b and c are formed by takingdifferences of the position vectors of these 3 points.Example 19Determine whether the points F (1, 1, 0) and G ( 8, 9, 14) lie onr (1 3t 3u ) i (2 t 3u ) j ( 1 3t 3u ) k .M Patel (April 2012)14St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesIf F lies on the plane then,1 3t 3u 12 t 3u 1 1 3t 3u 0Adding the first and third equations gives u 1/6. Substituting thisinto either the first or third equation then gives t 1/6. However,these solutions do not satisfy the second equation (check !). Thus, F doesnot lie on the plane.A similar analysis for G shows that u 1 and t 4 satisfy,1 3t 3u 82 t 3u 9 1 3t 3u 14Hence, G lies on the plane.Vector Equation of a Plane Given Normal and 1 Point on PlaneExample 20Determine a vector equation of the plane passing through the pointD (2, 3, 4) and which has normal vector 2 i 4 k .To find a vector equation, 2 vectors lying in the plane must be found. Thisis done by finding 2 vectors at right angles to the normal. Thus, 2solutions must be found to the equation, 2 p 0 q 0 4 r M Patel (April 2012)15St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes 2 0 By inspection, 2 solutions are 0 and 3 . There are infinitely many 1 0 solutions (draw a picture !).Hence, a vector equation for the plane is, 2 2 0 r 3 t 0 u 3 4 1 0 which can be written as,r (2 2t ) i (3 3u ) j (4 t ) kVector Equation of a Plane Given 2 Vectors in the PlaneExample 21Find a vector equation of the plane passing through W (1, 1, 1) and parallelto i 6 j and 5 i 3 j - k .A vector equation is, 1 1 5 r 1 t 6 u 3 1 0 1 r (1 t 5u ) i (1 6t 3u ) j (1 u ) kVector Equation of a Plane Given 3 Points in the PlaneExample 22Determine the equation of the plane containing the points A (1, 6, 0),B ( 4, 2, 5) and C ( 2, 4, 1).Two vectors lying in the plane are,M Patel (April 2012)16St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes 5 3 AB 8 and AC 10 5 1 Hence, a vector equation for the plane is, 1 5 3 r 6 t 8 u 10 0 5 1 r (1 5t 3u ) i ( 6 8t 10u ) j ( 5t u ) kConverting Between the Cartesian and Vector FormsExample 23Find the Cartesian equation of the plane which has vector equationr (3 t 2u) i (1 7t 4u) j (1 5t u) k .Rewriting the vector r gives, 3 1 2 r 1 t 7 u 4 1 5 1 which shows that the point (3, 1, 1) lies on the plane and that the vectors 1 2 7 and 4 lie in the plane. To find the Cartesian equation, we need a 5 1 point in the plane (which we have) and a normal that can be constructed inthe usual way, 13 1 2 n 7 x 4 9 10 5 1 M Patel (April 2012)17St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesThus, the Cartesian equation is, 13x 9y 10z 13 3 9 1 10 1 13x 9y 58 10z 58Example 24Find a vector equation for the plane which has Cartesian equation5x 2y 4z 13.A normal 5 to the plane is 2 . Two vectors at right angles to this normal 4 0 and 2 . Again, there are infinitely many choices for these 2 1 2 are 3 1 vectors. By inspection, the point (1, 2, 1) lies on the plane. A vectorequation for the plane is thus, 1 2 0 r 2 t 3 u 2 1 1 1 r (1 2t ) i ( 2 3t 2u ) j (1 t u ) kEquations of LinesEquations of lines can be determined in 2 ways, depending on theinformation given. 1 point on the line and a direction for the line. 2 points on the line.M Patel (April 2012)18St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesDefinition:A vector equation for a line in 3D with direction vector u a i b j c k passing through a point A( x 1 , y1 , z1 ) (with position vector a) and ageneral point P(x, y, z) with position vector p is,p a t u(t ℝ)Definition:The parametric form of a line, using the notation in the previousdefinition, is,x x 1 at,y y1 bt,z z1 ctDefinition:The symmetric form of a line (aka standard form or canonical form),using the notation in the previous definition (assuming none of a, b or cequal 0), is,x x1a y y1b z z1c tWhen at least one of a, b or c equal 0, we abandon the symmetric formand use the parametric form instead.It is easy to convert between the vector, parametric and symmetricforms of a line.As with planes, line equations are not unique.Example 25Show that the lines with equationsandx 21M Patel (April 2012) y2 z 13x 12 y 24 z 46are the same.19St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesLines are the same if their direction vectors are parallel and both lines 2 pass through the same point. The first line has direction 4 and the 6 1 second line has direction 2 . Combining the first line equation (with 3 parameter s) and the second line equation (with parameter t) gives oneequation (repeated thrice), namely, 2s 1 t. This equation hasinfinitely many solutions. In other words, infinitely many common pointslie on both lines. Hence, as the directions are parallel, the lines are thesame.Example 26Determine whether or not the points B ( 2, 7, 3) and C (1, 4, 1) lie ony 2x 1z 8the line .3511The parametric equations for this line are,x 1 3t,y 2 5t,z 8 11tSubstituting the coordinates of B into the parametric equations gives t 1 for each equation. Hence, as there is a unique t value, B lies on theline.Doing the same for C shows that t 0, t 6/5 and tthere is not a unique t value, C does not lie on the line. 7/11. AsEquation of a Line Given 1 Point on the Line and a DirectionExample 27Find the vector, parametric and symmetric forms of the line passingthrough T (4, 11, 17) and which is parallel to 3 i 5 j - 8 k .The vector equation is,M Patel (April 2012)20St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes x y z 4 11 17 3 t 5 8 The parametric equations are,x 4 3t,y 11 5t,z 17 8tThe symmetric form is,x 43y 11 5 z 17 8 t(t ℝ)Example 28Equation of a Line Given 2 Points on the LineFind the vector form of the line that passes through A (1, 4, 6) andB (3, 3, 3). 2 A direction vector for the line is 7 . The vector form of the line is 9 thus, x y z 1 4 6 2 t 7 9 Of course, this is just one of the vector forms.Example 29Find the value of k for which the linesandx 121 y 393 x 276 y 45 z 3z 2are perpendicular.kFor the lines to be perpendicular, the direction vectors of the lines musthave zero scalar product. Hence,M Patel (April 2012)21St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and Planes 3k 06 153k 21k 7Intersections of LinesAngle Between 2 LinesThe angle between 2 lines is the acute angle made by their directionvectors.Example 30Find the angle made by the linesx 121 y 393 x 272z 26 y 41 z 3and. 2 1 The directions of the lines are 1 and 3 . The scalar product 3 6 equations give, 13cos θ cos θ 0 · 512 . . .14 46θ 120 · 8 The acute angle between the lines is thus 59 · 2 .Definition:Lines in 3D are skew if they neither intersect nor are parallel.M Patel (April 2012)22St. Machar Academy
Advanced Higher Notes (Unit 3)Vectors, Lines and PlanesIn 2D, 2 lines can intersect in 1 point, intersect in infinitely many points(are coincident) or not intersect (are parallel).In 3D, the same situation occurs, except that in case of no intersection,the lines can be either parallel or skew.Intersection of 2 Lines in a PointExample 31Find the point of intersection of the lines x 5 (yy 3x 12z 5and .54 2 2) zParametrically, the lines take the respective forms,x 5 t,y 2 t,x 5s 12,y 3 2s ,z tz 5 4sEquating components, we find that,t 5s 7t 2s 1t 4s 5These are easily solved to give s 2 and t 3. The uniquesolution for s and t means that the lines intersect at one point only. Theintersection point is thus (2, 1, 3).Intersection of 2 Lines in a LineExample 32Show that the linesx 1210 M Patel (April 2012)x 12y 3 45 y 3 2z 58 z 54andintersec
Advanced Higher Notes (Unit 3) Vectors, Lines and Planes M Patel (April 2012) 1 St. Machar Academy Vectors, Lines and Planes Prerequisites: Adding, subtracting and scalar multiplying vectors; calculating angles between vectors. Maths Applications: Describing geometric transformations.
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The lines not intersecting and not parallel are called skew lines. Show L 0: . x 2s;y 3 s;z 3 4s are skew lines. Vectors and the Geometry of Space 23/29. The Dot Product The Cross Product Lines and Planes Lines Planes A plane in three dimensional space is determined by a point in the plane and a vector n perpendicular to the plane .
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