The angle between the straight lines $\overrightarrow{\mathrm{r}} =\left(2 - 3\mathrm{t}\right)\hat{\mathrm{i}} + \left(1 + 2\mathrm{t}\right)\hat{\mathrm{j}} + \left(2 + 6\mathrm{t}\right)\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{r}} =\left(1 + 4\mathrm{s}\right)\hat{\mathrm{i}} + \left(2 - s\right)\hat{\mathrm{j}} + \left(8s - 1\right)\hat{\mathrm{k}}$ is

• ${\cos }^{-1}\left(\frac{\sqrt{41}}{34}\right)$

• ${\cos }^{-1}\left(\frac{21}{34}\right)$

• ${\cos }^{-1}\left(\frac{43}{63}\right)$

• ${\cos }^{-1}\left(\frac{34}{63}\right)$

If Q is the image of the point P(2, 3, 4) under the reflection in the plane x - 2y + 5z = 6, then the equation of the line PQ is

• $\frac{\mathrm{x} - 2}{- 1} = \frac{\mathrm{y} - 3}{2} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{1} = \frac{\mathrm{y} - 3}{- 2} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{- 1} = \frac{\mathrm{y} - 3}{- 2} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{1} = \frac{\mathrm{y} - 3}{2} = \frac{\mathrm{z} - 4}{5}$

The distance of the point of intersection of the line $\frac{\mathrm{x} - 2}{3} = \frac{\mathrm{y} + 1}{4} = \frac{\mathrm{z} - 2}{12}$ and the plane x - y + z = 5 from the point (- 1, - 5, - 10)is

• 13

• 12

• 11

• 8

If the direction cosines of a line are $\left(\frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}\right)$, then

• 0 < c < 1

• c > 2

• $\mathrm{c} = \pm \sqrt{2}$

• $\mathrm{c} = \pm \sqrt{3}$

If the lines $\frac{1 - \mathrm{x}}{3} = \frac{\mathrm{y} - 2}{2\mathrm{\alpha }} = \frac{\mathrm{z} - 3}{2}$$\frac{\mathrm{x} - 1}{3\mathrm{\alpha }} = \mathrm{y} - 1 = \frac{6 - \mathrm{z}}{5}$ are perpendicular, then the value of $\mathrm{\alpha }$ is

• $\frac{- 10}{7}$

• $\frac{10}{7}$

• $\frac{- 10}{11}$

• $\frac{10}{11}$

The distance between the lines $\overrightarrow{\mathrm{r}} = \left(4\hat{\mathrm{i}} - 7\hat{\mathrm{j}} - 9\hat{\mathrm{k}}\right) + \mathrm{t}\left(3\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 4\hat{\mathrm{k}}\right)$ and $\overrightarrow{\mathrm{r}} = \left(7\hat{\mathrm{i}} - 14\hat{\mathrm{j}} - 5\hat{\mathrm{k}}\right) + \mathrm{s}\left(- 3\hat{\mathrm{i}} + 7\hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)$ is equal to

• 1

• $\frac{1}{2}$

• $\frac{3}{4}$

• 0

Let T_n denote the number of triangles which can equal to be formed by using the vertices of a regular polygon of n sides. If T_{n + 1} - T_n = 28, then n equals

• 4

• 5

• 6

• 8

A plane makes intercepts a, b, cat A, B, C on the coordinate axes respectively. If the centroid of the $∆$ABC is at (3, 2, 1), then the equation of the plane is

• x + 2y + 3z = 9

• 2x - 3y - 6z = 18

• 2x + 3y + 6z = 18

• 2x + y + 6z = 18

449.

If the plane 3x + y + 2z + 6 = 0 is parallel to the line $\frac{3\mathrm{x} - 1}{2\mathrm{b}} = 3 - \mathrm{y} = \frac{\mathrm{z} - 1}{\mathrm{a}}$, then the avlue of 3a + 3b is

• $\frac{1}{2}$

• $\frac{3}{2}$

• 3

• 4

The equation of the line passing through the point (3, 0,- 4) and perpendicular to the plane 2x - 3y + 5z - 7 = 0 is

• $\frac{\mathrm{x} - 2}{3} = \frac{\mathrm{y}}{- 3} = \frac{\mathrm{z} + 4}{5}$

• $\frac{\mathrm{x} - 3}{2} = \frac{\mathrm{y}}{- 3} = \frac{\mathrm{z} - 4}{5}$

• $\frac{\mathrm{x} - 2}{3} = \frac{- \mathrm{y}}{3} = \frac{\mathrm{z} + 4}{5}$

• $\frac{\mathrm{x} + 3}{2} = \frac{\mathrm{y}}{3} = \frac{\mathrm{z} - 4}{5}$