Physics · Motion In A Plane · NEET
The angle is exactly 90 degrees, so the vectors are perpendicular. Start from |A+B|^2 = |A-B|^2. Expand both sides: A^2 + B^2 + 2(A.B) = A^2 + B^2 - 2(A.B). Cancel A^2 and B^2. You get 4(A.B) = 0, so A.B = 0, which means AB cos(theta) = 0. Since A and B are non-zero, cos(theta) = 0 and theta = 90 degrees. This is a fixed result, not something you re-derive in the exam.
No. This is the most common mix-up. Equal magnitude of sum and difference does NOT mean the vectors are equal or that A = B. It only tells you the angle between them is 90 degrees. Their magnitudes A and B can be completely different (for example 3 and 7); as long as they meet at a right angle, |A+B| = |A-B| holds.
That is a different question. The vectors (A+B) and (A-B) are perpendicular to each other when A = B in magnitude (equal lengths), because (A+B).(A-B) = A^2 - B^2 = 0. Do not confuse this with the page's result: |A+B| = |A-B| means A is perpendicular to B (theta = 90). Read the question carefully to see which pair is being asked about.
Use the sign of A.B. If |A+B| > |A-B|, then A.B > 0, so the angle is acute (less than 90 degrees). If |A+B| < |A-B|, then A.B < 0, so the angle is obtuse (more than 90 degrees). Equality is the exact boundary at 90 degrees. This lets you quickly judge acute, right, or obtuse from the two magnitudes.
Write both magnitudes with the resultant formula. |A+B| = sqrt(A^2 + B^2 + 2AB cos(theta)) and |A-B| = sqrt(A^2 + B^2 - 2AB cos(theta)). Setting them equal and squaring, the only surviving term is 2AB cos(theta) = -2AB cos(theta), so 4AB cos(theta) = 0 and cos(theta) = 0. In under 15 seconds you reach theta = 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.
The vectors are perpendicular; the angle between them is 90 degrees. Memorise this directly, it appears in NEET/AIPMT almost verbatim.
Yes. Equal magnitude of the sum and the difference is exactly the condition A.B = 0, which is the definition of A perpendicular to B.
No. Only the angle is fixed at 90 degrees. The magnitudes A and B can be any non-zero values.
It tests whether you can turn a magnitude condition into a dot-product condition in one step. It rewards students who know A.B = 0 means perpendicular, which is the gateway to the dot product topic next.
When the vectors are perpendicular, |A+B| = |A-B| = sqrt(A^2 + B^2), the Pythagoras result. Both the sum and difference have the same length.
