Reflexion of the point $\left(\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }\right)$ in XY-plane is

• $\left(0, 0, \mathrm{\gamma }\right)$

• $\left(- \mathrm{\alpha }, - \mathrm{\beta }, \mathrm{\gamma }\right)$

• $\left(\mathrm{\alpha }, \mathrm{\beta }, - \mathrm{\gamma }\right)$

• $\left(\mathrm{\alpha }, \mathrm{\beta }, 0\right)$

The distance of the point (- 2, 4, - 5) from the line $\frac{\mathrm{x} + 3}{3} = \frac{\mathrm{y} - 4}{5} = \frac{\mathrm{z} + 8}{6}$ is

• $\sqrt{\frac{37}{10}}$

• $\frac{37}{10}$

• $\frac{\sqrt{37}}{10}$

• $\frac{37}{\sqrt{10}}$

The equation of straight line passing through the point (a, b, c) and parallel to Z-axis, is

• $\frac{\mathrm{x} - \mathrm{a}}{1} = \frac{\mathrm{y} - \mathrm{b}}{1} = \frac{\mathrm{z} - \mathrm{c}}{0}$

• $\frac{\mathrm{x} - \mathrm{a}}{0} = \frac{\mathrm{y} - \mathrm{b}}{1} = \frac{\mathrm{z} - \mathrm{c}}{1}$

• $\frac{\mathrm{x} - \mathrm{a}}{1} = \frac{\mathrm{y} - \mathrm{b}}{0} = \frac{\mathrm{z} - \mathrm{c}}{0}$

• $\frac{\mathrm{x} - \mathrm{a}}{0} = \frac{\mathrm{y} - \mathrm{b}}{0} = \frac{\mathrm{z} - \mathrm{c}}{1}$

If the equation of the locus of a point equidistant from the points (a₁, b₁) and (a₂, b₂) is (a₁ - a₂)r + (b₁ - b₂)y + c = 0, then the value of 'c' is

• $\frac{1}{2}\left({{\mathrm{a}}_2}^2 + {{\mathrm{b}}_2}^2 - {{\mathrm{a}}_1}^2 - {{\mathrm{b}}_1}^2\right)$

• ${{\mathrm{a}}_1}^2 - {{\mathrm{a}}_2}^2 + {{\mathrm{b}}_1}^2 - {{\mathrm{b}}_2}^2$

• $\frac{1}{2}\left({{\mathrm{a}}_1}^2 + {{\mathrm{a}}_2}^2 + {{\mathrm{b}}_1}^2 + {{\mathrm{b}}_2}^2\right)$

• $\sqrt{{{\mathrm{a}}_1}^2 + {{\mathrm{b}}_1}^2 - {{\mathrm{a}}_2}^2 - {{\mathrm{b}}_2}^2}$

A tetrahedron has vertices at 0(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(- 1, 1, 2). Then, the angle between the faces OAB and ABC will be

• ${\cos }^{-1}\left(\frac{19}{35}\right)$

• ${\cos }^{-1}\left(\frac{17}{31}\right)$

• $30°$

• $90°$

Distance between parallel planes 2x - 2y + z + 3 = 0 and 4x - 4y + 2z + 5 = 0, is

• $\frac{1}{2}$

• $\frac{1}{3}$

• $\frac{1}{4}$

• $\frac{1}{6}$

Given two vectors $\hat{\mathrm{i}} - \hat{\mathrm{j}} \mathrm{and} \hat{\mathrm{i}} + 2\hat{\mathrm{j}}$, the unit vector coplanar with the two vectors and perpendicular to first, is

• $\frac{1}{\sqrt{2}}\left(\hat{\mathrm{i}} + \hat{\mathrm{j}}\right)$

• $\frac{1}{\sqrt{5}}\left(2\hat{\mathrm{i}} + \hat{\mathrm{j}}\right)$

• $\pm \frac{1}{\sqrt{2}}\left(\hat{\mathrm{i}} + \hat{\mathrm{j}}\right)$

• None of these

If a and b are unit vectors and $\mathrm{\theta }$ is the angle between them, then the value of $\left|\cos \left(\frac{\mathrm{\theta }}{2}\right)\right|$ is

• $\frac{1}{2}\left|\mathrm{a} + \mathrm{b}\right|$

• $\frac{1}{2}\left|\mathrm{a} - \mathrm{b}\right|$

• $\frac{\left|\mathrm{a} - \mathrm{b}\right|}{\left|\mathrm{a} + \mathrm{b}\right|}$

• $\frac{\left|\mathrm{a} + \mathrm{b}\right|}{\left|\mathrm{a} - \mathrm{b}\right|}$

If $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& 1 + {\mathrm{a}}^3\\ \mathrm{b}& {\mathrm{b}}^2& 1 + {\mathrm{b}}^3\\ \mathrm{c}& {\mathrm{c}}^2& 1 + {\mathrm{c}}^3\end{array}\right|$ and vectors (1, a, a²), (1, b, b²) and (1, c, c²) are non-coplanar, then the product abc equals

• 2

• - 1

• 1

• 0

320.

If the length of perpendicular drawn from origin on a plane is 7 unit and its direction ratios are - 3, 2 and 6, then that plane is

• - 3x + 2y + 6z - 7 = 0

• - 3x + 2y + 6z - 49 = 0

• 3x - 2y + 6z + 7 = 0

• - 3x + 2y - 6z - 49 = 0