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.
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