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. \(Q=0\)

What is the torque of a force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$newton acting at a point $\overrightarrow{\mathrm{r}}=\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$ metre about the origin? (Given: $\overrightarrow{\mathrm{\tau }}<mspace\; width="1em"></mspace>=\overrightarrow{\mathrm{r}<mspace\; width="1em"></mspace>}\times \overrightarrow{\mathrm{F}}$)

1. $6\hat{\mathrm{i}}-6\hat{\mathrm{j}}+12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

2. $17\hat{\mathrm{i}}-6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

3. $-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

4. $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

Three non zero vectors $\overrightarrow{\mathrm{A}},<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}<mspace\; width="1em"></mspace>\&<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{C}}$ satisfy the relation $\overrightarrow{\mathrm{A}}·\overrightarrow{\mathrm{B}}=0<mspace\; width="1em"></mspace>\&<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{A}}·\overrightarrow{\mathrm{C}}=0$. Then $\overrightarrow{\mathrm{A}}$ can be parallel to:

(1)  $\overrightarrow{\mathrm{B}}$

(2)  $\overrightarrow{\mathrm{C}}$

(3)  $\overrightarrow{\mathrm{B}}·\overrightarrow{\mathrm{C}}$

(4)  $\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{C}}$

The scalar product of two vectors is 8 and the magnitude of vector product is $8\sqrt{3}$. The angle between them is:

(1)  $30°$

(2)  $60°$

(3)  $120°$

(4)  $150°$

Given: $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$. Out of the three vectors $\overrightarrow{\mathrm{a}},<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ two are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. The angles between the vectors are:

(A)  90$°$, 135$°$, 135$°$

(B)  30$°$, 60$°$, 90$°$

(C)  45$°$, 45$°$, 90$°$

(D)  45$°$, 60$°$, 90$°$

Vector $\overrightarrow{\mathrm{A}}$ is of length 2 cm and is 60$°$ above the x-axis in the first quadrant. Vector $\overrightarrow{\mathrm{B}}$ is of length 2 cm and 60$°$ below the x-axis in the fourth quadrant. The sum $\overrightarrow{\mathrm{A}}$+$\overrightarrow{\mathrm{B}}$ is a vector of magnitude -

(A)  2 along + y-axis

(B)  2 along + x-axis

(C)  1 along - x-axis

(D)  2 along - x-axis

Six forces, 9.81 N each, acting at a point are coplanar. If the angle between neighboring forces are equal, then the resultant is

(1)  0 N

(2)  9.81 N

(3)  2$\times$9.81 N

(4)  3$\times$9.81 N