Find the two unit vectors in R^2 parallel to the line y=3x+4.?
Hi, first of all I'd find a vector parallel to the line, and in order to do that you can either choose any two points P and Q on the line and consider the vector P - Q or write the line in parametric form x = t y = 3t + 4, where the coefficients of t are the components of a parallel vector. So you see that the line is parallel to the vector (1, 3). This vector has length l = sqrt(1^2 + 3^2) = sqrt(10), so a unit vector parallel to (1, 3) would be v = 1/sqrt(10) (1, 3) = (1/sqrt(10), 3/sqrt(10)) = (sqrt(10)/10, 3sqrt(10)/10). Once you have find one the other is just its opposite, that is w = -v = -(sqrt(10)/10, 3sqrt(10)/10) = (-sqrt(10)/10, -3sqrt(10)/10). I hope this helped you. Bye!
Find points on graph of z=3x^2 - 4y^2 at which the vector n=<3,2,2> is normal to tangent plane?
Let F(x,y,z) = 3x^2 - 4y^2 - z = 0. Its gradient <6x, -8y, -1> is normal to the surface. So, we need <6x, -8y, -1> to be directly proportional to <3, 2, 2>: <6x, -8y, -1> = k<3, 2, 2> for some k ==> 6x = 3k, -8y = 2k, and -1 = 2k. Since k = -1/2, we have x = -1/4 and y = 1/8. Substituting this into F, we find that z = 1/8. So, the only point is (-1/4, 1/8, 1/8). I hope this helps!
How do I find angle between line X-2/3=Y-3/5=Z-4/4 and plane 2x-2y+z+5=0?
Let a, b, c be the direction ratios of the line and A, B, C be the direction ratios of the normal to the plane... Let the angle between the line and the plane be m. Then angle between the normal to the plane and the line is 90-m (90 degrees). We know that the angle between two lines is given by :Cos (e) =(aA+bB+cC) /(L*N) where L=sqrt ((a^2) +(b^2)+(c^2)), N=sqrt ((A^2) +(B^2) +(C^2)) and 'e' is the angle between the lines. In this question, we need to find 'e' Put e=90 - m and substitute the value for a, b, c, A, B and C as per the assumption, find L and N, you will get the answer!! Hope this helps...
Find a vector which is parallel to the intersection of the planes 4x-3y+5z=12 and 2x+4y-z=-2?
Plane 4x - 3y + 5z = 12 is perpendicular to vector <4, -3, 5> Plane 2x + 4y - z = -2 is perpendicular to vector <2, 4, -1> Line of intersection of the two planes is perpendicular to both vectors To find direction vector of this line, simply take the cross product of the two vectors above: <4, -3, 5> x <2, 4, -1> = <-17, 14, 22> So vectors < -17, 14, 22 > and < 17, -14, -22 > and any vector that is a scalar multiple are parallel to the intersection of the planes.
Consider the vector field F(x,y,z)=(5z+y)i+(2z+x)j+(2y+5... Find a function f such that F=∇f and f(0,0,0)=0.?
Well, first of all, welcome to this forum from (maybe the only) french guy in the zone !! ;-) F(x,y,z)=(5z+y)i+(2z+x)j+(2y+5x)k. Necessary conditions: ∂f/∂x (x,y,z) = 5z+y integrating wrt x gives : there exists a function a(y,z) so that : f(x,y,z) = (5z + y)x + a(y,z) ∂f/∂y (x,y,z) = 2z+x integrating wrt y gives : there exists a function b(x,z) so that : f(x,y,z) = (2z+x)y + b(x,z) ∂f/∂z (x,y,z) = 2y+5x integrating wrt z gives : there exists a function c(x,y) so that : f(x,y,z) = (2y+5x)z + c(x,y) the identification brings : (5z + y)x + a(y,z) = (2z+x)y + b(x,z) = (2y+5x)z + c(x,y) a(y,z) = 2yz b(x,z) = 5xz c(x,y) = xy let's verify : f(x,y,z) = xy+2yz+5xz leads to : ∂f/∂x (x,y,z) = 5z+y ∂f/∂y (x,y,z) = 2z+x ∂f/∂z (x,y,z) = 2y+5x remark: the condition f(0,0,0)=0 is not needed because part of the necessary conditions, a possible constant is included in the functions a, b and c. conclusion : f(x,y,z) = xy+2yz+5xz only suitable vector field b) f is the potential of F therefore: ∫CF⋅dr = f(1,1,1) - f(0,0,0) = (1+2+5) - 0 = 8 et voilà, jeune homme !! ;-) hope it' ll help !! PS: if you want good answer, do not forget to give Best Answers..... to me, or anybody else, or course provided the answer deserves one ! this is to encourage people to answer !! ;-)
How do I get the vector equation of a line perpendicular to two other lines?
The first is going in direction (-1,2,1) [ignore the base point it is starting from] and the second in direction (1,1,-1). Take the cross-product (-2–1,1–1,-1–2)=(-3,0,-3) and that is the mutually perpendicular direction, or any scalar multiple like just (1,0,1). You want it to start from a point of intersection of the two lines, or if they’re skew you have to find their points of closest approach, but you’re in luck: -s=1+t so substitute s=-1-t into the others, 2-t=3–1-t, true, and -3+2t=1–1-t solves t=1, s=-2 and the point is (1,-1,2). So the line you want is x=1+r,y=-1,z=2+r
Sketching a curve and finding vector function?
x= t , y=t , and z(t) =2t^2 , so the curve is r(t) = t i+tj+2t^2 k z=x^2+y^2 is a paraboloid and y=x , z=z (any ) is a plane in the the line y=x , so z=2x^2 is a vertical parabola but in the plane y=x x^2+y^2 =9 i s a cylinder with a circle as a base , so it is really x^2+y^2 =9 , z=z ( any ) So , we have that at any z , x^2+y^2 =9 , x^2= 9-y^2 , plug , z=9- (9-y^2) z=y^2 If x=3cost , y^2 = 9-9cos ^2t = 9(1-cos^2t) =9sin^2t , so y=+-3sint z(t)= 9sin^2t r(t) = 3cost i+-3sint j +9sin^2t k x^2 = -(z+9) is vertical parabola in the plane ZX , vertex (0,-9) and opens downwards .- This is the base of a cylinder, so it is really x^2 = -(z+9) y=y ( any) x^2+y^2=9 , z=z (any) is a cylinder with base a circle R=3 The curve is really strange , but, sketch it .-( It is like a curved parabola opening downwards) x=3cost is only a parametrization .-