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The direction of the force between 3 point charges is always along the line connecting the two charges. If all three charges have the same sign, the force will be repulsive and if two charges have the same sign and one has the opposite sign, the force will be attractive. 3.
To find the equation of a plane given a point and a line, you need to first determine the normal vector of the plane. This can be done by taking the cross product of the direction vector of the line and a vector formed by subtracting the given point from any other point on the line. Once you have the normal vector, you can use the point-normal ...
So if we work out the equation of the line that goes through the point (1, 3, 6) which is perpendicular to the plane, then we can use it to find where it intersects the plane. Notice we already have the normal vector of the plane: (1, -1, 1), and so the line is l = (1, -1, 1)t + (1, 3, 6), where t takes on all real values.
For a infinite line of charge a distance R away, the electric field produced is E = 2kλ / R. For just a point charge some distance r away, the electric field is E = kq / r 2. The Attempt at a Solution I drew a picture in the x and y coordinate plane with the line of charge at x = -1, the point charge at (2, 3), and the analyze point at (3, 2.5).
To determine if three points lie on a straight line in Calc. 3, you can use the slope formula or the determinant method. If the slope between any two of the points is equal, then the points lie on a straight line. Alternatively, you can calculate the determinant of the matrix formed by the coordinates of the points.
The x-intercept is (-1/2, 0) creating the second point needed to find the slope. I will now solve the given equation for y. -2x + y = 1. y= 2x + 1. The equation I need to find can be found by finding the slope using (-1/2, 0) and (6, 2). I then plug one of the points and the slope into the point-slope formula.
Scalar Projection: Find Distance Point to Line. In summary, the scalar projection can be used to find the distance between a point (x1, y1) and a line ax + by + c = 0 by taking the scalar product of a unit normal to the line with a difference vector between the point and a point on the line. The unit normal can be found by taking the derivative ...
A point charge of 20 nC is located at (4,-1,-3), and a uniform line charge of -25nC/m lies along the intersection of the planes x=-4 and z=6. (a) Calculate D at (3,1,0). (b) How much electric flux leaves the surface of a sphere of radius 5, centered at the origin? (C) Repeat part b if the radius of the sphere is 10. Homework Equations
Problem: At a certain point in a horizontal pipeline, the water's speed is 2.50 m/s and the gauge pressure is 1.80 * 10^5 Pa. Find the gauge pressure at a second point in the line if the crosssection area at the second point is twice that of the first. Now I realize the critical part of this problem is finding the area, thus Pressure = Force/Area.
An implicit equation for the plane passing through the point (-2,5,-5) that is perpendicular to the line L(t) = <5+2t,3,4> is ...? Homework Equations a(x-x 0) + b(y-y 0) + c(z-z 0) = 0 The Attempt at a Solution So in order to find the equation of the plane I would need a normal vector and I got that using the following steps: