12.4 Cross Product
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
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
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 2 3 3 2 3 1 1 3 1 2 2 1 1 2 3 1 2 3 check: ( u v) u u v u v, u v u v, u v u v u, u, u similary: ( u v) v length... 2 3 3 2 3 1 1 3 1 2 2 1 1 2 3 u v u u v u u v u u v u u v u u v u 2 3 1 3 2 1 3 1 2 1 3 2 1 2 3 2 1 3
An easier way to remember the formula for the cross products is in terms of determinants: 2x2 determinant: a b ad bc c d 1 2 4 6 2 3 4 3x3 determinants: An example 1 6 2 3 1 3 4 5 2 Copy 1 st 2 columns 1 6 2 1 6 3 1 3 3 1 4 5 2 4 5 sum of forward diagonal products sum of backward diagonal products determinant = 2 72 3 36 15 8 4 59 19
i j k i j k i j u v u1 u2 u3 u1 u2 u3 u1 u2 v v v 1 2 3 v v v v v 1 2 3 1 2 iu v ju v ku v ku v iu v ju v 2 3 3 1 1 2 2 1 3 2 1 3 u v u v u v u v u v u v u v i j k 2 3 3 2 1 3 3 1 1 2 2 1 u v u v u v, u v u v, u v u v 2 3 3 2 3 1 1 3 1 2 2 1
Let u 1, 2,1 and v 3,1, 2 Find u v. i j k u v 1 2 1 3 1 2 i j k i j 1 2 1 1 2 3 1 2 3 1 u v 41 i 32 j 1 6 k uv 3,5,7
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
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 1 2 1 3 2 i j k i j 1 2 1 1 3 2 1 3 ( 6) i ( 2 2) j (3 ) k 6i 3k 6,,3 1 u v = 36 9 45 area = 45 2
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
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.
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 1 2 3 6,,3 6,,3 1,3, 4 6 12 6 ( u v) w 1 3 4 1 2 1 3 2 Volume = 6 6 12 6 6 6 i j k i j u u u u u 1 2 3 1 2 v v v v v 1 2 3 1 2 w w w 1 2 3 w1, w2, w3 = u1 u2 u3 = Triple scalar product v v v 1 2 3 (take absolute value)
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
12.5 Lines and Planes
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
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
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
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
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.
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, 5 1 2 3t 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 1 1 4 5 2 y 5 3 1 3 3 check Find the point of intersection using L1 : x z 3 1 1 4 1 4,2,3
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
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, 5 1 2 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 1 3 4 3 16 18 j 4 6 2 4 1 i k 13 7 2 i j k is normal or 13x 7y 2z 72 n 13,7,2 13( x 3) 7( y 5) 2( z 1)