Question

Vector A has a magnitude of 16 m and makes an angle of 44° with the positive x axis. Vector B also has a magnitude of 13 m and is directed along the negative x axis. Find A+B (in meters and degrees)

Find A-B (in meters and degrees)

Answer 1

Answer with explanation:

The given vectors in are reduced to their componednt form as shown

For vector A it can be written as

`\overrightarrow{v}_(a)=16cos(44^(o))\widehat{i}+16sin(44^(o))\widehat{j}`

Similarly vector B can be written as

`\overrightarrow{v}_(b)=-13\widehat{i}`

Hence The sum and difference is calculated as

`\overrightarrow{v}_(a)+\overrightarrow{v}_(b)=16cos(44^(o))\widehat{i}+16sin(44^(o))\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_(a)+\overrightarrow{v}_(b)=(16cos(44^(o))-13)\widehat{i}+16sin(44^(o))\widehat{j}\\\\\therefore \overrightarrow{v}_(a)+\overrightarrow{v}_(b)=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_(a)+\overrightarrow{v}_(b)|=\sqrt{(-1.49)^(2)+11.11^(2)}=11.21m`

The direction is given by

`\theta =tan^(-1)(r_(y))/(r_(x))\\\\\theta =tan^(-1)(11.11)/(-1.49)=97.64^(o)`with positive x axis.

Similarly

`\overrightarrow{v}_(a)-\overrightarrow{v}_(b)=16cos(44^(o))\widehat{i}+16sin(44^(o))\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_(a)-\overrightarrow{v}_(b)=(16cos(44^(o))+13)\widehat{i}+16sin(44^(o))\widehat{j}\\\\\therefore \overrightarrow{v}_(a)-\overrightarrow{v}_(b)=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_(a)-\overrightarrow{v}_(b)|=\sqrt{(24.51)^(2)+11.11^(2)}=26.91m`

The direction is given by

`\theta =tan^(-1)(r_(y))/(r_(x))\\\\\theta =tan^(-1)(11.11)/(24.51)=24.38^(o)`with positive x axis.