Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg \({\rm{B}}\), which is \(20.0\;{\rm{m}}\) in a direction exactly \(40^\circ \) south of west, and then leg \({\rm{A}}\), which is \(12.0\;{\rm{m}}\) in a direction exactly \(20^\circ \) west of north. (This problem shows that \({\rm{A}} + {\rm{B}} = {\rm{B}} + {\rm{A}}\).)

When two vectors are taken along two sides of a triangle, then the third side of the triangle is always in reverse order. The third side is said to be the resultant vector which is equal to the vector sum of the other two vectors.

The components of vectors are represented as,

Fig: Components of vector \({\rm{A}}\) and \({\rm{B}}\)

• The magnitude of vector \({\rm{A}}\)is, \(A = 12\;{\rm{m}}\).
• Angle vector \({\rm{A}}\) makes an angle, \({\theta _A} = {20^ \circ }\;{\rm{WN}}\).
• The magnitude of vector \({\rm{B}}\), \(B = 20\;{\rm{m}}\).
• Angle vector \({\rm{B}}\)makes an angle, \({\theta _B} = {40^ \circ }\;{\rm{SW}}\).

The horizontal component of the vector \({\rm{A}}\) is,

\({A_x} = - A\sin {\theta _A}\)

Here \(A\) is the magnitude of vector \({\rm{A}}\)and \({\theta _A}\) is the angle between \(y\)-axis and the vector \({\rm{A}}\).

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_x} &= - \left( {12\;{\rm{m}}} \right) \times \sin \left( {{{20}^ \circ }} \right)\\ &= - 4.1\;{\rm{m}}\end{aligned}\)

The horizontal component of the vector \({\rm{B}}\) is,

\({B_x} = - B\cos {\theta _B}\)

Here \(B\) is the magnitude of the vector \({\rm{B}}\), and \({\theta _B}\) is made by the vector \({\rm{B}}\) with the horizontal.

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{B_x} &= - \left( {20\;{\rm{m}}} \right) \times \cos \left( {{{40}^ \circ }} \right)\\ &= - 15.32\;{\rm{m}}\end{aligned}\)

The resultant horizontal component of the vector \({\rm{A}}\) and \({\rm{B}}\) is,

\({R_x} = {A_x} + {B_x}\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{R_x} &= \left( { - 4.1\;{\rm{m}}} \right) + \left( { - 15.32\;{\rm{m}}} \right)\\ &= - 19.42\;{\rm{m}}\end{aligned}\)

The vertical component of the vector \({\rm{A}}\) is,

\(\begin{aligned}{}{A_y} &= - \left( { - A} \right)\cos {\theta _A}\\ &= A\cos {\theta _A}\end{aligned}\)

Here \(A\) is the magnitude of vector \({\rm{A}}\)and \({\theta _A}\) is the angle between \(y\)-axis and the vector \({\rm{A}}\).

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_y} &= \left( {12\;{\rm{m}}} \right) \times \cos \left( {{{20}^ \circ }} \right)\\ &= 11.28\;{\rm{m}}\end{aligned}\)