question_answer

Two vectors \[\vec{A}\] and \[\vec{B}\]are of equal lengths (A = B) and mutually perpendicular. Show by vector diagram that their vector sum \[(\vec{A}+\,\vec{B})\] and vector difference \[(\vec{A}-\,\vec{B})\]will be of the same length and mutually perpendicular.

Answer:

                (i) The multiplication of a vector \[\vec{A}\]by a real number n becomes another vector \[n\vec{A}\]. Its magnitude becomes n times the magnitude of the given vector. Its direction is the same or opposite as that of\[\vec{A}\], according as n is a positive or negative real number, Thus  \[n\vec{A}=n\vec{A}\]and\[-n\vec{A}=-n\vec{A}\]. For example, if a vector \[\vec{A}\] is multiplied by a real number\[\text{n}=\text{2}\], we get \[2\vec{A}\], which is a vector, acting in the direction of \[\vec{A}\] and having magnitude twice as that of\[\vec{A}\], Fig. If vector \[\vec{A}\] is multiplied by real number\[\text{n}=-\text{2}\], then we get \[-\text{2}\vec{A}\], which is also a vector, acting in the opposite direction of \[\vec{A}\] and having magnitude twice as that of\[\vec{A}\], Fig The unit of \[\text{n}\vec{A}\]is the same as that of \[\vec{A}\]. (a)                                       (b) (ii) When a vector \[\vec{A}\] is multiplied by a scalar S, it becomes a vector S\[\vec{A}\], whose magnitude is S times the magnitude of \[\vec{A}\] and it acts along the direction of \[\vec{A}\]. The unit of S\[\vec{A}\] is different from the unit of vector \[\vec{A}\]. For illustration, if \[\vec{A}\] = 100 newton due west and S = 10 sec, then S\[\vec{A}\] = 10 second × 100 newton due west = 1000 newton ?second due west.