“Explain Triangle method (head to tail method) of vector addition.”
Let us consider two vectors $\vec{A}$ and $\vec{B}$ that lie in a plane as shown in figure $(a)$.
The lengths of the line segments representing these vectors are proportional to the magnitude of the vectors.
To find the sum $\vec{A}+\vec{B}$, we place vector $\vec{B}$ so that its tail is at the head of the vector $\vec{A}$, as in figure (b).
Then we join the tail of $\overrightarrow{\mathrm{A}}$ to the head of $\overrightarrow{\mathrm{B}}$.
This line $\overrightarrow{O Q}$ represent a vector $\vec{R}$, that is the sum of the vectors $\vec{A}$ and $\vec{B}$.
Since, in this procedure of vector addition, vectors are arranged head to tail, this graphical method is called the head-to-tail method.
The two vectors and their resultant form three sides of a triangle, so this method is also known as triangle method of vector addition.
Two forces $3\,N$ and $2\, N$ are at an angle $\theta$ such that the resultant is $R$. The first force is now increased to $ 6\,N$ and the resultant become $2R$. The value of is ....... $^o$
Two vectors $\dot{A}$ and $\dot{B}$ are defined as $\dot{A}=a \hat{i}$ and $\overrightarrow{\mathrm{B}}=\mathrm{a}(\cos \omega t \hat{\mathrm{i}}+\sin \omega t \hat{j}$ ), where a is a constant and $\omega=\pi / 6 \mathrm{rad} \mathrm{s}^{-1}$. If $|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{3}|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|$ at time $t=\tau$ for the first time, the value of $\tau$, in, seconds, is. . . . . .
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