91Ó°ÊÓ

When two vectors are taken along two sides of a triangle, the third side of the triangle is always in reverse order. It represents the magnitude and direction of the resultant vector, according to the triangle rule of vector addition.

• Magnitude of vector \({\rm{A}}\) \(A = 12\,{\rm{m}}\).
• Direction of vector \({\rm{A}}\) \({\theta _A} = 20^\circ \).
• Magnitude of vector \({\rm{B}}\) \(B = 20.0\,{\rm{m}}\).
• Direction of vector \({\rm{B}}\) \({\theta _B} = 40.0^\circ \).

The vectors are represented as,

Fig: Representation of vector

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

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

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

Substitute \(12{\rm{ m}}\) for \(A\) and \(20^\circ \) for \({\theta _A}\),

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

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 \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

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

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

\({R'_x} = {A_x} + {B_x}\)

Substitute \( - 4.1{\rm{ m}}\) for \({A_x}\) and \(15.32{\rm{ m}}\) for \({B_x}\),

\(\begin{aligned}{}{{R'}_x} = \left( { - 4.1{\rm{ m}}} \right) + \left( {15.32{\rm{ m}}} \right)\\ = 11.22{\rm{ m}}\end{aligned}\)

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

\({A_y} = A\cos {\theta _A}\)

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

Substitute \(12{\rm{ m}}\) for \(A\) and \(20^\circ \) for \({\theta _A}\),

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

The vertical component of the vector \(B\) is,

\({B_y} = B\sin {\theta _B}\)

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

Substitute \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

\(\begin{aligned}{}{B_y} = \left( {20{\rm{ m}}} \right) \times \sin \left( {40^\circ } \right)\\ = 12.86{\rm{ m}}\end{aligned}\)

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

\({R'_y} = {A_y} + {B_y}\)

Substitute \({\rm{11}}{\rm{.28 m}}\) for \({A_y}\) and \(12.86{\rm{ m}}\) for \({B_y}\),

\(\begin{aligned}{}{{R'}_y} = \left( {11.28{\rm{ m}}} \right) + \left( {12.86{\rm{ m}}} \right)\\ = 24.14{\rm{ m}}\end{aligned}\)

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

\[{R}'=\sqrt{{{R_x}'^2}+{{R_y}'^2}}\]

Substitute \(11.22{\rm{ m}}\) for \({R'_x}\) and \(24.14{\rm{ m}}\) for \({R'_y}\),

\(\begin{aligned}{}R' = \sqrt {{{\left( {11.22{\rm{ m}}} \right)}^2} + {{\left( {24.14{\rm{ m}}} \right)}^2}} \\ = 26.62{\rm{ m}}\end{aligned}\)

The direction of the resultant vector is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{{R'}_y}}}{{{{R'}_x}}}} \right)\)

Substitute \(11.22{\rm{ m}}\) for \({R'_x}\) and \(24.14{\rm{ m}}\) for \({R'_y}\),

\(\begin{aligned}{}\alpha = {\tan ^{ - 1}}\left( {\frac{{24.14{\rm{ m}}}}{{11.22{\rm{ m}}}}} \right)\\ = 65.1^\circ \end{aligned}\)

Hence, the magnitude of the resultant vector is \(26.62{\rm{ m}}\) and is directed towards \(65.1^\circ \) north of east.

The vectors are represented as,

Representation of vector

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

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

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

Substitute \(12{\rm{ m}}\) for \(A\) and \(20^\circ \) for \({\theta _A}\),

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

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 the angle made by the vector \({\rm{B}}\) with the horizontal.

Substitute \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

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

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

\({R''_x} = {A_x} + {B_x}\)

Substitute \(4.1{\rm{ m}}\) for \({A_x}\) and \( - 15.32{\rm{ m}}\) for \({B_x}\),

\(\begin{aligned}{}{{R''}_x} = \left( {4.1{\rm{ m}}} \right) + \left( { - 15.32{\rm{ m}}} \right)\\ = - 11.22{\rm{ m}}\end{aligned}\)

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

\({A_y} = - A\cos {\theta _A}\)

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

Substitute \(12{\rm{ m}}\) for \({\rm{A}}\) and \(20^\circ \) for \({\theta _A}\),

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

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

\({B_y} = - B\sin {\theta _B}\)

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

Substitute \(20{\rm{ m}}\) for \(B\) and \(40^\circ \) for \({\theta _B}\),

\(\begin{aligned}{}{B_y} = - \left( {20{\rm{ m}}} \right) \times \sin \left( {40^\circ } \right)\\ = - 12.86{\rm{ m}}\end{aligned}\)

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

\({R''_y} = {A_y} + {B_y}\)

Substitute \( - {\rm{11}}{\rm{.28 m}}\) for \({A_y}\) and \( - 12.86{\rm{ m}}\) for \({B_y}\),

\(\begin{aligned}{}{{R''}_y} = \left( { - 11.28{\rm{ m}}} \right) + \left( { - 12.86{\rm{ m}}} \right)\\ = - 24.14{\rm{ m}}\end{aligned}\)

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

\({R}'=\sqrt{{{R_x}'^2}+{{R_y}'^2}}\)

Substitute \( - 11.22{\rm{ m}}\) for \({R''_x}\) and \( - 24.14{\rm{ m}}\) for \({R''_y}\),

\(\begin{aligned}{}R'' = \sqrt {{{\left( {11.22{\rm{ m}}} \right)}^2} + {{\left( {24.14{\rm{ m}}} \right)}^2}} \\ = 26.62{\rm{ m}}\end{aligned}\)

The direction of the resultant vector is,

\(\beta = {\tan ^{ - 1}}\left( {\frac{{{{R''}_y}}}{{{{R''}_x}}}} \right)\)

Substitute \( - 11.22{\rm{ m}}\) for \({R''_x}\) and \( - 24.14{\rm{ m}}\) for \({R''_y}\),

\(\begin{aligned}{}\beta = {\tan ^{ - 1}}\left( {\frac{{ - 24.14{\rm{ m}}}}{{ - 11.22{\rm{ m}}}}} \right)\\ = 65.1^\circ \end{aligned}\)

Hence, the magnitude of the resultant vector is \(26.62{\rm{ m}}\) and is directed towards \(65.1^\circ \) south of west.