Geometric description of the cross product of the vectors u and v. The cross product of two vectors is a vector! u x v is perpendicular to u and v

Transcription

1 12.4 Cross Product

2 Geometric description of the cross product of the vectors u and v The cross product of two vectors is a vector! u x v is perpendicular to u and v The length of u x v is uv u v sin The direction is given by the right hand side rule

3 Right hand rule Place your 4 fingers in the direction of the first vector, curl them in the direction of the second vector, Your thumb will point in the direction of the cross product

4 Algebraic description of the cross product of the vectors u and v The cross product of u u, u, u and v v, v, v is u v u v u v, u v u v, u v u v check: ( u v) u u v u v, u v u v, u v u v u, u, u similary: ( u v) v length u v u u v u u v u u v u u v u u v u

5 An easier way to remember the formula for the cross products is in terms of determinants: 2x2 determinant: a b ad bc c d x3 determinants: An example Copy 1 st 2 columns sum of forward diagonal products sum of backward diagonal products determinant =

6 i j k i j k i j u v u1 u2 u3 u1 u2 u3 u1 u2 v v v v v v v v iu v ju v ku v ku v iu v ju v u v u v u v u v u v u v u v i j k u v u v u v, u v u v, u v u v

7 Let u 1, 2,1 and v 3,1, 2 Find u v. i j k u v i j k i j u v 41 i 32 j 1 6 k uv 3,5,7

8 Geometric Properties of the cross product: Let u and v be nonzero vectors and let be the angle between u and v. 1. u v is orthogonal to both u and v. 2. uv u v sin 3. uv if and only if u and v are scalar multiples of each other. u 4. uv area of the parallelogram u h u sin determined by u and v. v v 5. uv area of the triangle having 1 2 u u and v as adjacent sides. v

9 Problem: Compute the area of the triangle two of whose sides are given by the vectors u 1,,2 and v 1,3, 2 uv i j k i j k i j ( 6) i ( 2 2) j (3 ) k 6i 3k 6,,3 1 u v = area = 45 2

10 Algebraic Properties of the cross product: Let u, v, and w be vectors and let c be a scalar. 1. u v vu 2. uv w u v uw 3. c u v cu v u cv 4. v v 5. (c v)v 6. uvw u vw 7. uvw uw v u vw

11 Volume of the parallelepiped determined by the vectors a, b, and c. Area of the base bc Height comp a a cos b c a Volume bc a cos Volume a bc bc is called the scalar triple product this stands for absolute value The vectors are in the same plane coplanar if the scalar triple product is.

12 Problem: Compute the volume of the parallelepiped spanned by the 3 vectors u 1,,2, v 1,3, 2 and w 1,3, 4 Solution: From slide 9: Quicker: uv i j k ( u v) w u1 u2 u3 w1, w2, w3 v v v ,,3 6,,3 1,3, ( u v) w Volume = i j k i j u u u u u v v v v v w w w w1, w2, w3 = u1 u2 u3 = Triple scalar product v v v (take absolute value)

13 In physics, the cross product is used to measure torque. Q Consider a force F acting on a rigid body at a point given by a position vector r. The torque measures the tendency of the body to rotate about the origin point P rf r rf rfsin F is the angle between the force and position vectors

14 12.5 Lines and Planes

15 Recall how to describe lines in the plane (e.g. tangent lines to a graph): y mx b m is the slope b is the y intercept Point slope formula: y x y x m ( x, y ) is on the line Two point formula: y y y y x x x x 1 1 ( x, y ) and ( x, y ) 1 1 are on the line

16 Equations of Lines and Planes In order to find the equation of a line, we need : P x y z A a point on the line,, B a direction vector for the line v abc,, r r v t vector equation of line L L P x, y, z z P P tv r P x, y, z Here r is the vector from the origin to a specific point P on the line r is the vector from the origin to r v y a general point P ( x, y, z) on the line x r r x, y, z x, y, z P P tv t a, b, c v is a vector which is parallel to a vector that lies on the line v is not unique: 2 v, or v will also do

17 vector equation of the line L r r v or t x, y, z x, y, z t a, b, c equating components we get the parametric equations of the line L x x at, y y bt, z z ct Solving for t we get the symmetric equations of the line L x x y y z z a b c

18 Problem: P x y z A a point on the line,, B a direction vector for the line v abc,, Find the parametric equations of the line containing P 5,1,3 and P 3, 2,4. 1 choose P 5,1,3 v P P1 P1 P 3 5, 2 1, 4 3 2, 3,1 or v 2,3, 1 (could also choose P 3, 2, 4 ) x, y, z x, y, z t a, b, c 5,1,3 t 2,3, 1 The line is: x 5 2 t, y 1 3 t, z 3 t

19 Two lines in 3 space can interact in 3 ways: A) Parallel Lines - their direction vectors are scalar multiples of each other B) Intersecting Lines - there is a specific t and s, so that the lines share the same point. C) Skew Lines - their direction vectors are not parallel and there is no values of t and s that make the lines share the same point.

20 Problem: Determine whether the lines L and L are parallel, skew 1 2 or intersecting. If they intersect, find the point of intersection. L : x 3 t, y 5 3 t, z 1 4t : 8 2, 6 4, t 8 2s 53t 6 4s L x s y s z s Set the x coordinate equal to each other:, or 2s t 5 Set the y coordinate equal to each other:, or 4s 3t 11 We get a system of equations: 2s t 5 or 4s 3t 11 4s 2t 1 4s 3t 11 t 1 s 2 Check to make sure that the z values are equal for this t and s. 1 4t 5 s y check Find the point of intersection using L1 : x z ,2,3

21 Planes In order to find the equation of a plane, we need : P x y z A a point on the plane,, B a vector that is orthogonal to the plane n abc,, n r r nr nr vector equation of the plane this vector is called the normal vector to the plane n r r a x x b y y c z z r r OP x, y, z OP x, y, z n abc,, P P r - r x x, y y, z z scalar equation of the plane ax by cz ax by cz ax by cz d linear equation of the plane or

22 Problem: Determine the equation of the plane that contains the lines L and L. 1 2 L : x 3 t, y 5 3 t, z 1 4t : 8 2, 6 4, In order to find the equation of a plane, we need : A a point on the plane L x s y s z s can choose e.g. P (3,5, 1) B a vector that is orthogonal to the plane n abc,, We have two vectors in the plane: from L : u 1,3, 4 and from L : v 2, 4,1 1 2 i j k u v j i k i j k is normal or 13x 7y 2z 72 n 13,7,2 13( x 3) 7( y 5) 2( z 1)