draw a circle
vector PQ where P(-4, -3) and Q(-2, 2) equivalent vector P'Q' where P'(0, 0) and Q'(2, 5) the magnitude doesn't change so we can compute |P'Q'| = √(22 + 52) = √29
The answer depends on the information that you have about the four points and the manner in which that information is presented. Suppose the 4 points are A, B, C and D and the point that you find is P. If you have the coordinates of A, B, C and D then gradient AP = gradient AB (or any other pair) will suffice. If you have any one of vectors AB (or AC, AD, BC, BD), then vector AP is parallel to vector AB will suffice.
The Resultant Vector minus the other vector
We get the Unit Vector
If they are parallel, you can add them algebraically to get a resultant vector. Then you can resolve the resultant vector to obtain the vector components.
vector PQ where P(-4, -3) and Q(-2, 2) equivalent vector P'Q' where P'(0, 0) and Q'(2, 5) the magnitude doesn't change so we can compute |P'Q'| = √(22 + 52) = √29
The answer depends on the information that you have about the four points and the manner in which that information is presented. Suppose the 4 points are A, B, C and D and the point that you find is P. If you have the coordinates of A, B, C and D then gradient AP = gradient AB (or any other pair) will suffice. If you have any one of vectors AB (or AC, AD, BC, BD), then vector AP is parallel to vector AB will suffice.
The Resultant Vector minus the other vector
We get the Unit Vector
Divide the vector by it's length (magnitude).
If they are parallel, you can add them algebraically to get a resultant vector. Then you can resolve the resultant vector to obtain the vector components.
Determine the position of the object at two different times. The vector joining the first with the second of those positions points in the direction of the objects motion.
The component of a vector x perpendicular to the vector y is x*y*sin(A) where A is the angle between the two vectors.
find the vector<1,1>+<4,-3>
The normal vector to the surface is a radius at the point of interest.
1) The position vector of a particle is r= (a cosώt) i+ (a sinώt) j. The velocity of the particle is and find the parallel position vector.
Consider any two points on the vector, P = (a, b) and Q = (c, d). And lext x be the angle made by the vector with the positive direction of the x-axis. Then either a = c, so that the vector is vertical and its direction is straight up or a - c is non-zero. In that case, tan(x) = (b - d)/(a - c) or x = tan-1[(b - d)/(a - c)]