Please be patient as the PDF generation may take upto a minute.

Two forces A and B have a resultant $R_1$. If B is doubled, the new resultant $R_2$ is perpendicular to A. Then
1. $R_1=A$
2. $R_1=B$
3. $R_2=A$
4. $R_2=B$

The value of unit vector in the direction of vector $\mathrm{A}=5 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}$, is -
1. $\hat{\mathrm{i}}$
2. $\hat{\mathrm{j}}$
3. $\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)/13$
4. $\left(5\hat{\mathrm{i}}-12\hat{\mathrm{j}}\right)/13$

The linear velocity of a rotating body is given by $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{\omega }}\times \overrightarrow{\mathrm{r}}$, where $\mathrm{\omega }$ is the angular velocity and r is the radius vector. The angular velocity of a body, $\overrightarrow{\mathrm{\omega }}=\hat{\mathrm{i}}-2\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and their radius vector is  $\overrightarrow{\mathrm{r}}=4\hat{\mathrm{j}}-3\hat{\mathrm{k}}, \mathrm{then\; value\; of} \left|\overrightarrow{\mathrm{v}}\right|$ will be:
1. $\sqrt{29} \mathrm{units}$
2. $31 \mathrm{units}$
3. $\sqrt{37} \mathrm{units}$
4. $\sqrt{41} \mathrm{units}$

The vector having a magnitude of 10 and perpendicular to the vector $3\hat{\mathrm{i}}-4\hat{\mathrm{j}}$ is- 
1. $4\hat{\mathrm{i}}-3\hat{\mathrm{j}}$
2. $5\sqrt{2} \hat{\mathrm{i}}-5\sqrt{2} \hat{\mathrm{j}}$
3. $8\hat{\mathrm{i}}+6\hat{\mathrm{j}}$
4.  $8\hat{\mathrm{i}}-6\hat{\mathrm{j}}$

If a, b and c are three non-zero vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$, then the value of $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}}$ will be:

The angle made by the vector $\overrightarrow{\mathrm{A}} = \sqrt{3}\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ with y-axis is:
1.  ${\sin }^{-1}\left(\frac{3}{\sqrt{14}}\right)$
2.  ${\sin }^{-1}\left(\frac{\sqrt{7}}{4}\right)$
3.  ${\cos }^{-1}\left(\frac{4}{3}\right)$
4.  ${\cos }^{-1}\left(\frac{3}{5}\right)$

\(\overrightarrow A\) and \(\overrightarrow {B}\) are two vectors given by \(\overrightarrow {A}= 2\hat i + 3\hat j\) and \(\overrightarrow {B}= \hat i + \hat j\). The component of \(\overrightarrow A\) parallel to \(\overrightarrow B\) is:
1. \(\frac{(2\hat i -\hat j)}{2}\)
2. \(\frac{5}{2}(\hat i - \hat j)\)
3. \(\frac{5}{2}(\hat i + \hat j)\)
4. \(\frac{(3\hat i -2\hat j)}{2}\)

If $\overrightarrow{\mathrm{A}}=4\hat{\mathrm{i}}-3\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{B}}=6\hat{\mathrm{i}}+8\hat{\mathrm{j}}$ then magnitude and direction of $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ will be
1. $5, {\tan }^{-1}(3/4)$
2. $5\sqrt{5}, {\tan }^{-1}(1/2)$
3. $10, {\tan }^{-1}\left(5\right)$
4. $25, {\tan }^{-1}(3/4)$