If the vectors $\overrightarrow{\mathrm{a}} = \hat{\mathrm{i}} +\hspace{0.17em}\mathrm{a}\hat{\mathrm{j}} +\hspace{0.17em}{\mathrm{a}}^2\hat{\mathrm{k}}$, $\overrightarrow{\mathrm{b}}= \hat{\mathrm{i}} +\hspace{0.17em}\mathrm{b}\hat{\mathrm{j}} +\hspace{0.17em}{\mathrm{b}}^2\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}} = \hat{\mathrm{i}} +\hspace{0.17em}\mathrm{c}\hat{\mathrm{j}} +\hspace{0.17em}{\mathrm{c}}^2\hat{\mathrm{k}}$ are three non-coplanar vectors and $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& 1 + {\mathrm{a}}^3\\ \mathrm{b}& {\mathrm{b}}^2& 1 + {\mathrm{b}}^3\\ \mathrm{c}& {\mathrm{c}}^2& 1 + {\mathrm{c}}^3\end{array}\right| = 0$, then the value of abc is

• 0

• 1

• 2

• - 1

Let $\overrightarrow{\mathrm{a}} = 2\hat{\mathrm{i}} - \hat{\mathrm{j}} + \hat{\mathrm{k}}$, $\overrightarrow{\mathrm{b}} = \hat{\mathrm{i}} + 2\hat{\mathrm{j}} - \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}} = \hat{\mathrm{i}} + \hat{\mathrm{j}} - 2\hat{\mathrm{k}}$  be three vectors. A vector in the plane of $\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{c}}$ whose projection on $\overrightarrow{\mathrm{a}}$ is magnitude $\sqrt{\frac{2}{3}}$, is

• $2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} - 3\hat{\mathrm{k}}$

• $2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 3\hat{\mathrm{k}}$

• $2\hat{\mathrm{i}} - 5\hat{\mathrm{j}} + 5\hat{\mathrm{k}}$

• $2\hat{\mathrm{i}} + \hat{\mathrm{j}} + 5\hat{\mathrm{k}}$

If the constant forces $\mathrm{}$ 2\hat{\mathrm{i}} - 5\hat{\mathrm{j}} + 6\hat{\mathrm{k}}$\stackrel{}{}$ and $$$- \hat{\mathrm{i}} + 2\hat{\mathrm{j}} - \hat{\mathrm{k}}$act on a particle due to which it is displaced from a point A (4,- 3, - 2) to a point B (6, 1,- 3), then the work done by the forces is

• 15 unit

• 9 unit

• - 15 unit

• - 9 unit

If $\overrightarrow{\mathrm{a}} . \hat{\mathrm{i}} = 4, \mathrm{then} \left(\overrightarrow{\mathrm{a}} \times \hat{\mathrm{j}}\right) . \left(2\hat{\mathrm{j}} - 3\hat{\mathrm{k}}\right)$ is equal to

• 12

• 2

• 0

• - 12

$\overrightarrow{\mathrm{a}} \times \left[\overrightarrow{\mathrm{a}} \times \left(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right)\right]$ is equal to

• $\left(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}\right) . \left(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}\right)$

• $\overrightarrow{\mathrm{a}} . \left(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}\right) - \overrightarrow{\mathrm{b}}\left(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right)$

• $\left[\overrightarrow{\mathrm{a}} . \left(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right)\right]\overrightarrow{\mathrm{a}}$

• $\left(\overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{a}}\right) \left(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}\right)$

If the vectors $\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$, $- \hat{\mathrm{i}} + 2\hat{\mathrm{j}}$ represents the diagonals of a parallelogram, then its area will be

• $\sqrt{21}$

• $\frac{\sqrt{21}}{2}$

• $2\sqrt{21}$

• $\frac{\sqrt{21}}{4}$

The position vector of the points A, B, C are $\left(2\hat{\mathrm{i}} + \hat{\mathrm{j}} - \hat{\mathrm{k}}\right)$, $\left(3\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + \hat{\mathrm{k}}\right)$ and $\left(\hat{\mathrm{i}} + 4\hat{\mathrm{j}} - 3\hat{\mathrm{k}}\right)$ respectively. These points

• form an isosceles triangle

• form a right angled triangle

• are collinear

• form a scalene triangle

538.

If $\left|\overrightarrow{\mathrm{a}}\right| = 2, \left|\overrightarrow{\mathrm{b}}\right| = 3$ and $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are mutually perpendicular, then the area of the triangle whose vertices are $\overrightarrow{0}, \overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}}$ is

• 5

• 1

• 6

• 8

Given $\overrightarrow{\mathrm{p}} = 3\hat{\mathrm{i}} +\hspace{0.17em}2\hat{\mathrm{j}} + 4\hat{\mathrm{k}}$, $\overrightarrow{\mathrm{a}} = \hat{\mathrm{i}} +\hspace{0.17em}\hat{\mathrm{j}}$, $\overrightarrow{\mathrm{c}} = \hat{\mathrm{i}} +\hspace{0.17em}\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{p}} = \mathrm{x}\overrightarrow{\mathrm{a}} + \mathrm{y}\overrightarrow{\mathrm{b}} + \mathrm{z}\overrightarrow{\mathrm{c}}$ then x, y, z are respectively

• $\frac{3}{2}, \frac{1}{2}, \frac{5}{2}$

• $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}$

• $\frac{5}{2}, \frac{3}{2}, \frac{1}{2}$

• $\frac{1}{2}, \frac{5}{2}, \frac{3}{2}$

If $2\overrightarrow{\mathrm{a}} + 3\overrightarrow{\mathrm{b}} - 5\overrightarrow{\mathrm{c}} = \overrightarrow{0}$, then ratio in which $\overrightarrow{\mathrm{c}}$ divides $\overrightarrow{\mathrm{AB}}$ is

• 3 : 2 internally

• 3 : 2 externally

• 2 : 3 internally

• 2 : 3 externally