Match List$- I$ with List$- II.$
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Choose the correct answer from the options given below :
$(a) \rightarrow(iv),(b) \rightarrow(i),(c) \rightarrow(i i i),(d) \rightarrow(i i)$
$(a) \rightarrow(iv),(b) \rightarrow(iii),(c) \rightarrow(i),(d) \rightarrow(ii)$
$(a) \rightarrow(iii),(b) \rightarrow(ii),(c) \rightarrow(iv),(d) \rightarrow(i)$
$(a) \rightarrow(i),(b) \rightarrow(iv),(c) \rightarrow(ii),(d) \rightarrow(iii)$
A particle is simultaneously acted by two forces equal to $4\, N$ and $3 \,N$. The net force on the particle is
$ABC$ is an equilateral triangle. Length of each side is $a$ and centroid is point $O$. then $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=.......$
There are two force vectors, one of $5\, N$ and other of $12\, N $ at what angle the two vectors be added to get resultant vector of $17\, N, 7\, N $ and $13 \,N$ respectively
The position vectors of points $A, B, C$ and $D$ are $\vec A = 3\hat i + 4\hat j + 5\hat k,\,\vec B = 4\hat i + 5\hat j + 6\hat k,\,\vec C = 7\hat i + 9\hat j + 3\hat k$ and $\vec D = 4\hat i + 6\hat j$ then the displacement vectors $\overrightarrow {AB} $ and $\overrightarrow {CD} $ are
The resultant of two vectors $\overrightarrow P $ and $\overrightarrow Q $ is $\overrightarrow R .$ If $Q$ is doubled, the new resultant is perpendicular to $P$. Then $R $ equals