Coordinates of a moving particle are given by $\mathrm{x}={\mathrm{ct}}^2$ and $\mathrm{y}={\mathrm{bt}}^2$. The speed of the particle is given by

(1)  $2\mathrm{t} \left(\mathrm{c}+\mathrm{b}\right)$

(2)  $2\mathrm{t} \sqrt{{\mathrm{c}}^2-{\mathrm{b}}^2}$

(3)  $\mathrm{t} \sqrt{{\mathrm{c}}^2+{\mathrm{b}}^2}$

(4)  $2\mathrm{t} \sqrt{{\mathrm{c}}^2+{\mathrm{b}}^2}$

The x and y components of vector $\overrightarrow{\mathrm{A}}$ are 4m and 6m respectively. The x, y components of vector $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ are 10m and 9m respectively. The length of $\overrightarrow{\mathrm{B}}$ is ______ and angle that $\overrightarrow{\mathrm{B}}$ makes with the x axis is given by _______.

(A)  $\sqrt{0.45}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

(B)  $\sqrt{4.5}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

(C)  $\sqrt{45}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

(D)  $4\sqrt{5}\mathrm{m}, {\tan }^{-1}\left(1/2\right)$

A particle travels with speed 50 m/s from the point (3, 7) in a direction $7\hat{\mathrm{i}}-24\hat{\mathrm{j}}$. Find its position vector after 3 seconds.

1.  $151\hat{\mathrm{i}}+45\hat{\mathrm{j}}$

2.  $45\hat{\mathrm{i}}-137\hat{\mathrm{j}}$

3.  $151\hat{\mathrm{i}}-45\hat{\mathrm{j}}$

4.  $4.5\hat{\mathrm{i}}-151\hat{\mathrm{j}}$

If \(\overrightarrow{A}={2}\hat{i}+\hat{j}\;{\&}\;\overrightarrow{B}=\hat{i}{-}\hat{j},\)  then the components of \(\overrightarrow {A}\) along with \(\overrightarrow {B}\) & perpendicular to \(\overrightarrow {B}\) respectively will be:
1. \(\frac{\hat{i} - \hat{j}}{2} ,~\frac{3}{2}\left(\hat i+\hat j\right)\)
2. \(\frac{\hat{i} - \hat{j}}{2} ,~-\frac{2}{3}\left(\hat i+\hat j\right)\)
3. \(\frac{\hat{i} - \hat{j}}{2} ,~-\frac{3}{2}\left(\hat i-\hat j\right)\)
4. \(\frac{\hat{i} - \hat{j}}{2} ,~\frac{2}{3}\left(\hat i-\hat j\right)\)

If \(\overrightarrow{a}\) is a vector and \(x\) is a non-zero scalar, then which of the following is correct?

If \(\theta\) is the angle between two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$, and $|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}$, then \(\theta\) is equal to:
1. \(0^\circ\)
2. \(180^\circ\)
3. \(135^\circ\)
4. \(45^\circ\)

The vector \(\overrightarrow b\) which is collinear with the vector \(\overrightarrow a = \left(2, 1, -1\right)\) and satisfies the condition \(\overrightarrow a. \overrightarrow b=3\) is:
1. \(\left(1, \frac{1}{2}, \frac{-1}{2}\right)\)
2. \(\left(\frac{2}{3}, \frac{1}{3}, \frac{-1}{3}\right)\)
3. \(\left(\frac{1}{2}, \frac{1}{4}, \frac{-1}{4}\right)\)
4. \(\left(1, 1, 0\right)\)

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:

Two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ inclined at an angle $\mathrm{\theta }$ with respect to each other have a resultant $\overrightarrow{\mathrm{c}}$ which makes an angle $\mathrm{\beta }$ with $\overrightarrow{\mathrm{a}}$. If the directions of $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ are interchanged, then the resultant will have the same

(A)  magnitude

(B)  direction

(C)  magnitude as well as direction

(D)  neither magnitude nor direction

Two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ lie in a plane. Another vector $\overrightarrow{\mathrm{C}}$ lies outside this plane. The resultant $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}$ of these three vectors

(1)  can be zero

(2)  cannot be zero

(3)  lies in the plane of $\overrightarrow{\mathrm{A}} \&\overrightarrow{\mathrm{B}}$

(4)  lies in the plane of $\overrightarrow{\mathrm{A}} \& \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$