The angle between the pair of lines $\frac{\mathrm{x} - 2}{2} = \frac{\mathrm{y} - 1}{5} = \frac{\mathrm{z} + 3}{- 3}$ and $\frac{\mathrm{x} + 2}{- 1} = \frac{\mathrm{y} - 4}{8} = \frac{\mathrm{z} - 5}{4}$ is

• ${\cos }^{-1}\left(\frac{21}{9\sqrt{38}}\right)$

• ${\cos }^{-1}\left(\frac{23}{9\sqrt{38}}\right)$

• ${\cos }^{-1}\left(\frac{24}{9\sqrt{38}}\right)$

• ${\cos }^{-1}\left(\frac{26}{9\sqrt{38}}\right)$

The equation of the plane passing through the intersection of the planes x + 2y + 3z + 4 = 0 and 4x + 3y + 2z + 1 = 0 and the origin is :

• 3x + 2y + z + 1 = 0

• 3x + 2y + z = 0

• 2x + 3y + z = 0

• x + y + z = 0

If $\left(\frac{1}{2}, \frac{1}{3}, \mathrm{n}\right)$ are the direction cosines of a line, then the value of n is

• $\frac{\sqrt{23}}{6}$

• $\frac{23}{36}$

• $\frac{2}{3}$

• $\frac{3}{2}$

514.

The equation of the plane passing through (2, 3, 4) and parallel to the plane 5x - 6y + 7z = 3 is :

• 5x - 6y + 7z + 20 = 0

• 5x - 6y + 7z - 20 = 0

• - 5x + 6y - 7z + 3 = 0

• 5x + 6y + 7z + 3 = 0

The inclination of the straight line passing through the point (- 3, 6) and the mid point of the line joining the points ( 4, - 5) and (- 2, 9) is:

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{3\mathrm{\pi }}{4}$

A point moves such that the area of the triangle formed by it with the points (1, 5) and (3, - 7) is 21 sq unit Then locus of the point is :

• 6x + y - 32 = 0

• 6x - y + 32 = 0

• x + 6y - 32 = 0

• 6x - y - 32 = 0

The line $\frac{\mathrm{x}}{\mathrm{a}} - \frac{\mathrm{y}}{\mathrm{b}}$ = 1 cuts the x-axis at P. The equation of the line through P perpendicular to the given line is :

• x + y = ab

• x + y = a + b

• ax + by = a²

• bx + ay = b²

The value of $\mathrm{\lambda }$. for which the lines 3x + 4y = 5, 2x + 3y = 4and $\mathrm{\lambda }$x + 4y = 6 meet at a point, is :

• 2

• 1

• 4

• 3

Three vertices of a parallelogram taken in order are (- 1, -  6), (2, - 5) and (7, 2). The fourth vertex is :

• (1, 4)

• (1, 1)

• (4, 4)

• (4, 1)

The orthocentre of the triangle whose vertices are (5, - 2), - 1, 2) and (1, 4), is :

• $\left(- \frac{1}{5}, \frac{16}{5}\right)$

• $\left(\frac{14}{5}, \frac{1}{5}\right)$

• $\left(\frac{1}{5}, \frac{1}{5}\right)$

• $\left(\frac{14}{5}, \frac{14}{5}\right)$