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

632.

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

If direction cosines of a vector of magnitude 3 are $\frac{2}{3}, - \frac{9}{3}, \frac{{\displaystyle 2}}{{\displaystyle 3}}$, then vector is

• $\hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$

• $2\hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}}$

• $\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$

• None of these

Equation of line passing through the point (2, 3, 1) and parallel to the line of intersection of the planes x - 2y - z + 5 = 0 and x + y + 3z = 6 is

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

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

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

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

Foot of perpendicular drawn from the origin to the plane 2x - 3y + 4z = 29 is

• (7, - 1, 3)

• (5, - 1, 4)

• (5, - 2, 3)

• (2, - 3, 4)

Suppose $\mathrm{a} + \mathrm{b} + \mathrm{c} = 0, \left|\mathrm{a}\right| = 3, \left|\mathrm{b}\right| = 5, \left|\mathrm{c}\right| = 7$, then the angle between a and b is

• $\mathrm{\pi }$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{4}$

The vector equation of the plane, which is at a distance of $\frac{3}{\sqrt{14}}$, from the origin and the normal from the origin is $2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + \hat{\mathrm{k}}$ is

• $\mathrm{r} . \left(2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + \hat{\mathrm{k}}\right) = 3$

• $\mathrm{r} . \left(\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}\right) = 9$

• $\mathrm{r} . \left(\hat{\mathrm{i}} + 2\hat{\mathrm{j}}\right) = 3$

• $\mathrm{r} . \left(2\hat{\mathrm{i}} + \hat{\mathrm{k}}\right) = 3$

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 5y + 8 = 0.

• $\left(0, - \frac{18}{5}, 2\right)$

• $\left(0, \frac{8}{5}, 2\right)$

• $\left(\frac{8}{25}, 0, 0\right)$

• $\left(0, - \frac{8}{5}, 0\right)$