Physics · Motion In A Plane · NEET
Use the dot (scalar) product. Since A.B = AB cos(theta), rearrange to get cos(theta) = (A.B) / (AB). Then theta = cos-inverse of [ (A.B) / (AB) ]. Here A.B is the dot product and A, B are the magnitudes of the two vectors. This one formula works for any two vectors in 2D or 3D.
First take the dot product component by component: if A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k, then A.B = Ax*Bx + Ay*By + Az*Bz. Next find the magnitudes A = sqrt(Ax^2 + Ay^2 + Az^2) and B the same way. Finally cos(theta) = (A.B)/(AB). Example: A = 3i + 4j, B = 4i - 3j gives A.B = 12 - 12 = 0, so theta = 90 degrees.
Because the DOT product carries the cosine: A.B = AB cos(theta). The dot product is a scalar and is largest when vectors point the same way (theta = 0). The CROSS product carries the sine: |A x B| = AB sin(theta). So if a question gives or asks for a dot product, use cos; if it involves a cross product (like torque or area), use sin. For finding the plain angle, dot product with cos is the simplest route.
Yes. A negative dot product means the angle is obtuse (between 90 and 180 degrees), because cos(theta) is negative there. A zero dot product means theta = 90 degrees (perpendicular). A positive dot product means theta is acute (below 90 degrees). So just plug the signed value into theta = cos-inverse of [ (A.B)/(AB) ] and you get the correct angle.
This is a common NEET shortcut. Square both sides: |A+B|^2 = |A-B|^2. This gives A^2 + B^2 + 2A.B = A^2 + B^2 - 2A.B, so 4 A.B = 0, meaning A.B = 0. A zero dot product means the vectors are perpendicular, so theta = 90 degrees. Whenever sum and difference have equal magnitude, the vectors are at 90 degrees.
If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
theta = cos-inverse of [ (A.B) / (AB) ], where A.B is the dot product and A, B are the magnitudes of the vectors.
Compute A.B = Ax*Bx + Ay*By + Az*Bz, find magnitudes A and B, then use cos(theta) = (A.B)/(AB) and take cos-inverse.
When their dot product A.B = 0. A quick test: if |A + B| = |A - B|, the vectors are perpendicular, so theta = 90 degrees.
No. The dot product is commutative (A.B = B.A), so the angle between A and B is the same as between B and A.
The dot product directly contains cos(theta) as a scalar (A.B = AB cos theta), so dividing by the magnitudes isolates the angle cleanly. The cross product gives sin(theta) and a direction, which is more work for a plain angle.
