Question

the resultant of three vectors 1,2and 3units,whose directions are those of the side of an equilateral triangle,is at an angle of
A 30 deg with1
b 15 deg with 1
c. 100 deg with 1
d. 150 deg with 1

Answer 1

The vectors are acting along as the sides of an equilateral triangle as in first fig.

Here $\left|\vec{\sf{A}}\right|=\mathsf{1,\ }\ \left|\vec{\sf{B}}\right|=\mathsf{2}$ and $\left|\vec{\sf{C}}\right|=\mathsf{3.}$

The second fig. shows how if the three vectors were concurrent. From this fig. we get,

$\longrightarrow\vec{\sf{A}}=\sf{\hat i}$

$\longrightarrow\vec{\sf{B}}=\sf{2\cos120^o\,\hat i+2\sin120^o\,\^j}$

$\longrightarrow\vec{\sf{C}}=\sf{3\cos(-120^o)\,\hat i+3\sin(-120^o)\,\^j}$

So the resultant vector is,

$\longrightarrow\vec{\sf{R}}=\vec{\sf{A}}+\vec{\sf{B}}+\vec{\sf{C}}$

$\longrightarrow\vec{\sf{R}}=\sf{\hat i+2\cos120^o\,\hat i+2\sin120^o\,\^j+3\cos(-120)^o\,\hat i+3\sin(-120)^o\,\^j}$

$\longrightarrow\vec{\sf{R}}=\sf{(1+2\cos120^o+3\cos(-120^o))\,\hat i+(2\sin120^o+3\sin(-120^o))\,\^j}$

Since $\sf{\cos(-120^o)=\cos120^o}$ and $\sf{\sin(-120^o)=-\sin120^o,}$

$\longrightarrow\vec{\sf{R}}=\sf{(1+2\cos120^o+3\cos120^o)\,\hat i+(2\sin120^o-3\sin120^o)\,\^j}$

$\longrightarrow\vec{\sf{R}}=\sf{\left(1-2\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{2}\right)\,\hat i+\left(2\cdot\dfrac{\sqrt3}{2}-3\cdot\dfrac{\sqrt3}{2}\right)\,\^j}$

$\longrightarrow\vec{\sf{R}}=\sf{-\dfrac{3}{2}\,\hat i-\dfrac{\sqrt3}{2}\,\^j}$

The resultant vector makes the same angle with the vector A as that made with positive X axis, since A is acting along positive X axis.

Let $\theta$ be the angle between them, which is given by,

$\longrightarrow\theta=\cos^{-1}\left(\dfrac{\vec{\sf{A}}\cdot\vec{\sf{R}}}{\left|\vec{\sf{A}}\right|\cdot\left|\vec{\sf{R}}\right|}\right)$

$\longrightarrow\theta=\cos^{-1}\left(\dfrac{\sf{\left(-\dfrac{3}{2}\,\hat i-\dfrac{\sqrt3}{2}\,\^j\right)\,\hat i}}{\sf{\sqrt{\left(-\dfrac{3}{2}\right)^2+\left(-\dfrac{\sqrt3}{2}\right)^2}}}\right)$

$\longrightarrow\theta=\cos^{-1}\left(\dfrac{\sf{\left(-\dfrac{3}{2}\right)}}{\sf{\sqrt3}}\right)$

$\longrightarrow\theta=\cos^{-1}\sf{\left(-\dfrac{\sqrt3}{2}\right)}$

Since $\sf{0^o\ \textless\ \theta\ \textless\ 180^o,}$

$\longrightarrow\underline{\underline{\sf{\theta=150^o}}}$

Hence the resultant is at an angle of 150° with the vector of magnitude 1 unit.