Muhammad Muttahar Ghaffari Pirzada on Nearpeer MDCAT Common Room
Plz explain this question below 👇 @sir Mansoor
Alisha Nazar · 2026-06-19T17:50:17.791Z
A.B=ABcos¢, = ABcos(90), A.B= 0, because cos90° is equal to 0 . € this sign show teta. I think ye question is tarha solve huga.
Muhammad Muttahar Ghaffari Pirzada · 2026-06-19T18:34:19.021Z
More explanation plz
Tamkanat · 2026-06-19T17:56:34.232Z
B
Muhammad Muttahar Ghaffari Pirzada · 2026-06-19T18:34:09.352Z
How
Alisha Nazar · 2026-06-19T18:41:18.048Z
Ask from sir tomorrow inshallah
Ali Farooqi · 2026-06-19T20:18:32.641Z
it must be B. cz angle is less than 90 and greater than 0
Void Whisperer · 2026-06-19T17:47:10.081Z
There's a mistake; if your question is considered correct, answer is (B) & if your answer (A) is considered correct, it should have had "R.A=0"
Muhammad Muttahar Ghaffari Pirzada · 2026-06-19T18:34:02.573Z
Correct is b mine is wrong that's i asked how to solve this
Void Whisperer · 2026-06-19T18:58:13.998Z
As R is a perpendicular to A, B is the hypotenuse of the right-angled triangle, forming an acute angle at the head of A. When the triangle is opened & B is put tail-to-tail with A, the angle between the two vectors becomes obtuse (>90°) & cos(90° - 180°) = -ve