For three vectors p, q and r,if r = 3p + 4q a | vedclass

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(B) Given equations are:$r = 3p + 4q$  $(i)$$2r = p - 3q$  $(ii)$From equation $(ii)$,we have $p = 2r + 3q$.Substituting this value of $p$ in equation

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$|r| > 2|q|$ and $r, q$ have opposite directions

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(B) Given equations are:$r = 3p + 4q$  $(i)$$2r = p - 3q$  $(ii)$From equation $(ii)$,we have $p = 2r + 3q$.Substituting this value of $p$ in equation $(i)$:$r = 3(2r + 3q) + 4q$$r = 6r + 9q + 4q$$r - 6r = 13q$$-5r = 13q$$r = -\frac{13}{5}q$Since the scalar multiplier is negative,$r$ and $q$ have opposite directions.Taking the magnitude on both sides:$|r| = |-\frac{13}{5}q| = \frac{13}{5}|q| = 2.6|q|$Since $2.6 > 2$,we have $|r| > 2|q|$.Thus,$|r| > 2|q|$ and $r, q$ have opposite directions.

For three vectors $p, q$ and $r$,if $r = 3p + 4q$ and $2r = p - 3q$,then

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