If three numbers are drawn at random successively without replacement from a set S = {1, 2, ... 10}, then the probability that the minimum of the chosen numbers is 3 or their maximum is 7

• $\frac{11}{40}$

• $\frac{5}{40}$

• $\frac{3}{40}$

• $\frac{1}{40}$

$$\begin{gathered} \mathrm{If} \mathrm{the} \mathrm{points} \mathrm{having} \mathrm{the} \mathrm{position} \mathrm{vectors} \\ 3\hat{\mathrm{i}} - 2\hat{\mathrm{j}} - \hat{\mathrm{k}}, 2\mathrm{i} + 3\hat{\mathrm{j}} - 4\hat{\mathrm{k}}, - \hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}} \mathrm{and} \\ 4\hat{\mathrm{i}} + 5\hat{\mathrm{j}} + \mathrm{\lambda }\hat{\mathrm{k}} \mathrm{are} \mathrm{coplanar}, \mathrm{then} \mathrm{\lambda } = ? \end{gathered}$$

• $- \frac{146}{17}$

• 8

•  - 8

• $\frac{146}{17}$

If a, b and c are non-zero vectors such that a and b are not perpendicular to each other, then the vector r which is perpendicular to a and satisfying r x b = c x b  is

• $\frac{\left(\mathrm{a} \times \mathrm{b}\right) \times \mathrm{c}}{\mathrm{c} \mathrm{a}}$

• $\frac{\mathrm{b} \times \left(\mathrm{a} \times \mathrm{c}\right)}{\mathrm{b} \mathrm{c}}$

• $\frac{\left(\mathrm{b} \times \mathrm{c}\right) \times \mathrm{a}}{\mathrm{a} \mathrm{b}}$

• $\frac{\left(\mathrm{c} \times \mathrm{b}\right) \times \mathrm{a}}{\mathrm{ac}}$

$$\begin{gathered} \mathrm{The} \mathrm{triad} (\mathrm{x}, \mathrm{y}, \mathrm{z}) \mathrm{of} \mathrm{real} \mathrm{number} \mathrm{such} \mathrm{that} \\ \left(3\hat{\mathrm{i}} - \hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right) = \left(2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} - \hat{\mathrm{k}}\right)\mathrm{x} + \left(\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)\mathrm{y} + \left( - 2\hat{\mathrm{i}} + \hat{\mathrm{j}} - 2\hat{\mathrm{k}}\right)\mathrm{z} \mathrm{is} \end{gathered}$$

• (- 2, 5, 3)

• (2, - 5, 3)

• (2, 5, 3)

• (2, 5, - 3)

485.

If the volume of the tetrahedron formed by the coterminous edges a, b and c is 4, then the volume of the parallelopiped formed by the coterminous edges a x b, b x c and c x a is

• 576

• 48

• 16

• 144

$$\begin{gathered} \mathrm{Let} \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{c}} \mathrm{be} \mathrm{three} \mathrm{unit} \mathrm{vectors} \mathrm{such} \mathrm{that} {\left|\overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{c}}\right|}^2 + {\left|\overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{c}}\right|}^2 = 8. \\ \mathrm{Then} {\left|\overrightarrow{\mathrm{a}} + 2\overrightarrow{\mathrm{b}}\right|}^2 + {\left|\overrightarrow{\mathrm{a}} + 2\overrightarrow{\mathrm{c}}\right|}^2 = ? \end{gathered}$$

Let the position vectors of points 'A' and 'B' be $\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}} \mathrm{and} 2\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}$, respectively. A point 'P' divides the line segment AB internally in the ratio $\mathrm{\lambda }$ : 1 ($\mathrm{\lambda }$ > 0). If O is the origin and $\overrightarrow{\mathrm{OB}} . \overrightarrow{\mathrm{OP}} - 3\overrightarrow{\mathrm{OA}} \times {\overrightarrow{\mathrm{OP}}}^2 = 6$ then $\mathrm{\lambda }$ is equal to

$$\begin{gathered} \mathrm{Let} \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R} \mathrm{be} \mathrm{such} \mathrm{that} {\mathrm{a}}^2 + {\mathrm{b}}^2 + {\mathrm{c}}^2 = 1. \mathrm{If} \\ \mathrm{acos}\left(\mathrm{\theta }\right) = \mathrm{bcos}\left(\mathrm{\theta } + \frac{2\mathrm{\pi }}{3}\right) = \mathrm{ccos}\left(\mathrm{\theta } + \frac{4\mathrm{\pi }}{3}\right), \mathrm{where} \mathrm{\theta } = \frac{\mathrm{\pi }}{9}, \\ \mathrm{then} \mathrm{the} \mathrm{angle} \mathrm{between} \mathrm{the} \mathrm{vectors} \mathrm{a}\hat{\mathrm{i}} + \mathrm{b}\hat{\mathrm{j}} + \mathrm{c}\hat{\mathrm{k}} \mathrm{and} \mathrm{b}\hat{\mathrm{i}} + \mathrm{c}\hat{\mathrm{j}} +\hspace{0.17em}\hat{\mathrm{k}} \mathrm{is}: \end{gathered}$$

• 0

• $\frac{2\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{9}$

Let x₀ be the point of local maxima of $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \overrightarrow{\mathrm{a}} . \left(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}\right), \mathrm{where} \overrightarrow{\mathrm{a}} = \mathrm{x}\hat{\mathrm{i}} - 2\hat{\mathrm{j}} +\hspace{0.17em}3\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}} = - 2\hat{\mathrm{i}} + \mathrm{x}\hat{\mathrm{j}} - \hat{\mathrm{k}} \mathrm{and} \\ \overrightarrow{\mathrm{c}} = 7\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + \mathrm{x}\hat{\mathrm{k}}. \mathrm{Then} \mathrm{the} \mathrm{value} \mathrm{of} \overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{b}} + \overrightarrow{\mathrm{b}} . \overrightarrow{\mathrm{c}} + \overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}} \mathrm{at} \mathrm{x} = {\mathrm{x}}_0 \end{gathered}$$

is :

• - 22

• 14

•  - 4

•  - 30

Suppose the vectors x₁, x₂ and x₃ are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b₁, b₂ and b₃ respectively. If

${\mathrm{x}}_1 = \left[\begin{array}{c}1\\ 1\\ 1\end{array}\right], {\mathrm{x}}_2 = \left[\begin{array}{c}0\\ 2\\ 1\end{array}\right], {\mathrm{x}}_3 = \left[\begin{array}{c}0\\ 0\\ 1\end{array}\right], {\mathrm{b}}_1 = \left[\begin{array}{c}1\\ 0\\ 0\end{array}\right], {\mathrm{b}}_2 = \left[\begin{array}{c}0\\ 2\\ 0\end{array}\right]$,

then the determinant of A is equal of A is

• $\frac{3}{2}$

• 4

• $\frac{1}{2}$

• 2