Yes, provided it is the ray. If AB is a vector then the answer is no.
To find the resultant of 2 vectors, P and Q, let the ray AB represent the vector P. Let AB (not BA) be in the direction of P and let the length of AB represent the magnitude of P. Let BC represent the direction of Q and the length BC represent the magnitude of Q [on the same scale used for P and AB]. Then the straight line AC, which is the diagonal of the parallelogram with sides representing P and Q, is the resultant vector R, with magnitude and direction represented by AC.The vectors P and Q can also be represented as sides AB and AC. In that case you will need to complete the parallelogram and the resultant is represented by the diagonal through A.
Yes.
yes
[(aa + bb) + (ab+ba)(aa+bb)*(ab+ba)]*[a + (ab+ba)(aa+bb)*b]
Yes, provided it is the ray. If AB is a vector then the answer is no.
To find the resultant of 2 vectors, P and Q, let the ray AB represent the vector P. Let AB (not BA) be in the direction of P and let the length of AB represent the magnitude of P. Let BC represent the direction of Q and the length BC represent the magnitude of Q [on the same scale used for P and AB]. Then the straight line AC, which is the diagonal of the parallelogram with sides representing P and Q, is the resultant vector R, with magnitude and direction represented by AC.The vectors P and Q can also be represented as sides AB and AC. In that case you will need to complete the parallelogram and the resultant is represented by the diagonal through A.
Yes.
yes it is
yes
If these are vectors, then ba = - ab
The GCF is ab
According to the symmetric property (and common sense) line segmetn AB is congruet to line segment BA since they are the same segment, just with a different name
[(aa + bb) + (ab+ba)(aa+bb)*(ab+ba)]*[a + (ab+ba)(aa+bb)*b]
A.....................B ------------->----- | | | | | | ^ | | | C With the limited editing capability, draw a line AC equal to 3 units, The first vector to seattle is represented by the vector CA (not AC). Draw a line AB 2.5 units long and perpendicular to CA at A. The second vector is AB (not BA). Now join CB. The line CB (not BC) is the resultant vector both in direction and magnitude. Disregard the dots between A and B. The editor will not allow me to put a bunch of spaces between A and B.
NB, Nb
Line BA