A vector `vec A` of magniude (A) is turned through an angle ` theta`. Calculate the cahng in the magniude of vector.

Explanation

` vec A-1` and vec A-2` be the initial and final vector, where
` vec A_1 = A= A-2`. Find 2 (c ) . 96, then change in vector
` | Delta vec A| = | vec A-2 - vec A-1 | = | vec A_2 + ( -vec A|1)|`
.
Since the angle between ` vec A_1` and vec A_2` is ` theta`,
therefore angle between ` vec A_2` and ` (-vev A-1) is
` (180^@- theta)`.
:. Change in magnitude `=Delta A`
` = sqrt( A_1^2 + A_2^2 + 2 A_2 A_2 cos ( 190^@ - theta))`
` = sqrt(A^2 + A^2 - 2 AA cos theta)`
` = sqrt( 2 A^2 -2 A^2 cos theta ) = sqrt( 2 A^2 ( 1- cos theta )`
` = sqrt ( 2 A^2 xx 2 sin^2 theta//2)=2 A sin theta//2` .