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 $\overrightarrow{\mathrm{A}} \& \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$

The vector sum of the forces of 10 N and 6 N can be

(1)  2 N

(2)  8 N

(3)  18 N

(4)  20 N

A set of vectors taken in a given order gives a closed polygon. Then the resultant of these vectors is a 

(1)  scalar quantity

(2)  pseudovector

(3)  unit vector

(4)  null vector

The vector sum of two vectors P and Q is minimum when the angle $\mathrm{\theta }$ between their positive directions, is

(A)  $\frac{\mathrm{\pi }}{4}$

(B)  $\frac{\mathrm{\pi }}{3}$

(C)  $\frac{\mathrm{\pi }}{2}$

(D)  $\mathrm{\pi }$

If the vector sum of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is maximum, then the angle $\mathrm{\theta }$ between two vectors will be:

1.  $0°$

2.  $30°$

3.  $45°$

4.  $60°$

If $\stackrel{}{\mathrm{}}$\(\overrightarrow P +\overrightarrow Q\) = \(\overrightarrow P-\overrightarrow Q\) and \(\theta\) is the angle between \(\overrightarrow {P}\) and \(\overrightarrow {Q}\), then:
1. \(\theta = 0^{\circ}\)
2. \(\theta = 90^{\circ}\)
3. \(P=0\)
4. \(Q=0\)