If the vectors $3\hat{\mathrm{i}} + \hat{\mathrm{j}} - 2\hat{\mathrm{k}}$, $\hat{\mathrm{i}} + 2\hat{\mathrm{j}} - 3\hat{\mathrm{k}}$, $3\hat{\mathrm{i}} + \mathrm{\lambda }\hat{\mathrm{j}} + 5\hat{\mathrm{k}}$ are co-planar, the value of $\mathrm{\lambda }$ is

• - 4

• 4

• 8

• - 8

A space vector makes the angles 150° and 60° with the positive direction of x-and y-axes. The angle made by the vector with the positive direction z-axis is

• 90°

• 60°

• 180°

• 120°

If a, b and c are non-coplanar, then the value of $\mathrm{a} . \left\{\frac{\mathrm{b} \times \mathrm{c}}{3\mathrm{b} . \left(\mathrm{c} \times \mathrm{a}\right)}\right\} - \mathrm{b} . \left\{\frac{\mathrm{c} \times \mathrm{a}}{2\mathrm{c} . \left(\mathrm{a} \times \mathrm{b}\right)}\right\}$ is

• $- \frac{1}{2}$

• $- \frac{1}{3}$

• $- \frac{1}{6}$

• $\frac{1}{6}$

If ${\sin }^{-1}\left(\mathrm{a}\right)$ is the acute angle between the curves : x² + y² = 4x and x² + y² = 8 at (2, 2), then a is equal to

• 1

• 0

• $\frac{1}{\sqrt{2}}$

• $\frac{\sqrt{3}}{2}$

The distance of the point P(a, b, c) from the x-axis is

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

• $\sqrt{{\mathrm{b}}^2 + {\mathrm{c}}^2}$

• a

• $\sqrt{{\mathrm{a}}^2 + {\mathrm{c}}^2}$

Equation of the plane perpendicular to the line $\frac{\mathrm{x}}{1} = \frac{\mathrm{y}}{2} = \frac{\mathrm{z}}{3}$ and passing through the point (2, 3, 4) is

• 2x + 3y + z = 17

• x + 2y + 3z = 9

• 3x + 2y + z = 16

• x + 2y + 3z = 20

297.

The line $\frac{\mathrm{x} - 2}{3} = \frac{\mathrm{y} - 3}{4} = \frac{\mathrm{z} - 4}{5}$ is parallel to the plane

• 2x + 3y + 4z = 0

• 3x + 4y + 5z = 7

• 2x + y - 2z = 0

• x + y + z = 2

Te angle between two diagonals of a cube is

• ${\cos }^{-1}\left(\frac{1}{3}\right)$

• $30°$

• ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

• $45°$

Lines $\frac{\mathrm{x} - 2}{1} = \frac{\mathrm{y} - 3}{1} = \frac{\mathrm{z} - 4}{- \mathrm{k}}$ and $\frac{\mathrm{x} - 1}{\mathrm{k}} = \frac{\mathrm{y} - 4}{2} = \frac{\mathrm{z} - 5}{1}$ are coplanar, if

• k = 2

• k = 0

• k = 3

• k = - 1

If a = $\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}, \left|\mathrm{b}\right| = 5$ and the angle between a and b is $\frac{\mathrm{\pi }}{6}$, then the area of the triangle formed by these two vectors as two sides is

• $\frac{15}{4}$

• $\frac{15}{2}$

• $\frac{15\sqrt{3}}{2}$

• 15