The velocity of a boat X relative to a boat Y is $5\hat{\mathrm{i}} - 2\hat{\mathrm{j}}$ and that of Y relative to another boat Z is $9\hat{\mathrm{i}} + 4\hat{\mathrm{j}}$ where $\hat{\mathrm{i}} \mathrm{and} \hat{\mathrm{j}}$ are the velocity of k not per hour, east and north respectively. Then the velocity is :

• $\frac{\sqrt{2}}{10} \mathrm{knot}/\mathrm{h}$

• $\frac{10}{\sqrt{2}}$ knot/h

• $10\sqrt{2}$ knot/h

• $2\sqrt{10}$ knot/h

42.

Two vectors $\overrightarrow{\mathrm{a}} \mathrm{and} \overrightarrow{\mathrm{b}}$ of equal magnitude 5 originating from a point and directs respectively towards north-east and north-west. Then the magnitude of $\overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}}$ is :

• $3\sqrt{2}$

• $2\sqrt{3}$

• $2\sqrt{5}$

• $5\sqrt{2}$

The shortest distance from the point (1, 2, - 1) to the surface of the sphere x² + y² + z² = 24 is :

• $3\sqrt{6}$ unit

• $\sqrt{6}$ unit

• $2\sqrt{6}$

• 2 sq unit

ABCD is a quadrilateral, P, Q are the mid points of $\overrightarrow{\mathrm{BC}} \mathrm{and} \overrightarrow{\mathrm{AD}}$, then $\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{DC}}$ is equal to :

• 3$\overrightarrow{\mathrm{QP}}$

• $\overrightarrow{\mathrm{QP}}$

• $4\overrightarrow{\mathrm{QP}}$

• $2\overrightarrow{\mathrm{QP}}$

The equation of the plane which bisects the line joining (2, 3, 4) and (6, 7, 8) is :

• x - y - z - 15 = 0

• x - y - z - 15 = 0

• x + y + z - 15 = 0

• x + y + z + 15 = 0

If $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are are the three vectors mutually perpendicular to each other and $\left|\overrightarrow{\mathrm{a}}\right| = 1, \left|\overrightarrow{\mathrm{b}}\right| = 3 \mathrm{and} \left|\overrightarrow{\mathrm{c}}\right| = 5$, then $\left|\overrightarrow{\mathrm{a}} - 2\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{b}} - 3\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{c}} - 4\overrightarrow{\mathrm{a}}\right|$ is equal to

• 0

• - 24

• 3600

• - 215

A line makes acute angles of $\mathrm{\alpha }, \mathrm{\beta } \mathrm{and} \mathrm{\gamma }$ with the co-ordinate axes such that $\cos \left(\mathrm{\alpha }\right)\cos \left(\mathrm{\beta }\right) = \cos \left(\mathrm{\beta }\right)\cos \left(\mathrm{\gamma }\right) = \frac{2}{9} \mathrm{and} \cos \left(\mathrm{\gamma }\right)\cos \left(\mathrm{\alpha }\right) = \frac{4}{9}$, then $\cos \left(\mathrm{\alpha }\right) + \cos \left(\mathrm{\beta }\right) + \cos \left(\mathrm{\gamma }\right)$ is equal to :

• $\frac{25}{9}$

• $\frac{5}{9}$

• $\frac{5}{3}$

• $\frac{2}{3}$

A unit vector coplanar with $\hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}} \mathrm{and} \hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \hat{\mathrm{k}}$, and perpendicular to $\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}$ is :

• $\left(\frac{\hat{\mathrm{j}} - \hat{\mathrm{k}}}{\sqrt{2}}\right)$

• $\left(\frac{\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}}{\sqrt{3}}\right)$

• $\left(\frac{\hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}}}{\sqrt{6}}\right)$

• $\left(\frac{\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \hat{\mathrm{k}}}{\sqrt{6}}\right)$

Forces acting on a particle have magnitude 5, 3 and 1 unit and act in the direction of the vectors $6\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}, 3\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 6\hat{\mathrm{k}} \mathrm{and} 2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} - 6\hat{\mathrm{k}}$ respectively. They remain constant while the particle is displaced from the point A (2, - 1, - 3) to B (5, - 1, 1). The work done is :

• 11 unit

• 33 unit

• 10 unit

• 30 unit

The equation of the plane through the point (1, 2, 3), (- 1,  4, 2) and (3, 1, 1) is :

• 5x + y + 12z - 23 = 0

• 5x + 6y + 2z - 23 = 0

• x + 6y + 2z - 13 = 0

• x + y + z - 13 = 0