Chapter 3 Problem 8

Problem

Let ${\vec {V}}_{1}=-6.0{\hat {i}}+8.0{\hat {j}}$ and ${\vec {V}}_{2}=4.5{\hat {i}}-5.0{\hat {j}}$. Determine the magnitude and direction of

(a) ${\vec {V}}_{1}$

(b) ${\vec {V}}_{2}$

(c) ${\vec {V}}_{1}+{\vec {V}}_{2}$

(d) ${\vec {V}}_{2}-{\vec {V}}_{1}$

Solution

(a)

${\vec {V}}_{1}=-6.0{\hat {i}}+8.0{\hat {j}}$

$V_{1}={\sqrt {(-6.0)^{2}+(8.0)^{2}}}=10$

$\theta =\tan ^{-1}{\frac {8.0}{-6.0}}=127^{\circ }$

(b)

${\vec {V}}_{2}=4.5{\hat {i}}-5.0{\hat {j}}$

$V_{2}={\sqrt {(4.5)^{2}+(-5.0)^{2}}}=6.7$

$\theta =\tan ^{-1}{\frac {-5.0}{4.5}}=312^{\circ }$

(c)

${\vec {V}}_{1}+{\vec {V}}_{2}=(-6.0{\hat {i}}+8.0{\hat {j}})+(4.5{\hat {i}}-5.0{\hat {j}})=-1.5{\hat {i}}+3.0{\hat {j}}$

$\left|{\vec {V}}_{1}+{\vec {V}}_{2}\right|={\sqrt {(-1.5)^{2}+(3.0)^{2}}}=3.4$

$\theta =\tan ^{-1}{\frac {3.0}{-1.5}}=117^{\circ }$

(d)

${\vec {V}}_{2}-{\vec {V}}_{1}=(4.5{\hat {i}}-5.0{\hat {j}})-(-6.0{\hat {i}}+8.0{\hat {j}})=10.5{\hat {i}}-13.0{\hat {j}}$

$\left|{\vec {V}}_{2}-{\vec {V}}_{1}\right|={\sqrt {(10.5)^{2}+(-13.0)^{2}}}=16.7$

$\theta =\tan ^{-1}{\frac {-13.0}{10.5}}=309^{\circ }$