The Resultant Vector minus the other vector
If you assume the vector is only in two dimensions, you can find the missing y-component with Pythagoras' Theorem: y = square root of (magnitude2 - x2).
We get the Unit Vector
Missing....? But the answer to find the missing is normally either "algebra" or "find another equation to use."
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.
The Resultant Vector minus the other vector
If you assume the vector is only in two dimensions, you can find the missing y-component with Pythagoras' Theorem: y = square root of (magnitude2 - x2).
We get the Unit Vector
Divide the vector by it's length (magnitude).
Missing....? But the answer to find the missing is normally either "algebra" or "find another equation to use."
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.
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.
R3 is a complete vector room, so you can actually take *ANY* other vector, e.g. from r1, r2 or r4 or any other vector room.
You can find a missing denominator if you know something that the fraction is equal to. Then you can find the missing denominator through cross multiplication.