Mathematics : Three Dimensional Geometry

The point where the line $\frac{\mathrm{x} - 1}{2} = \frac{\mathrm{y} - 2}{- 3} = \frac{\mathrm{z} + 3}{4}$ meets the plane 2x + 4y - z = 1, is

• (3, - 1, 1)

• (3, 1, 1)

• (1, 1, 3)

• (1, 3, 1)

A vector vis equally inclined to the x-axis, y-axis and z-axis respectively, its direction cosines are

• $< \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} >$

• $< - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}} >$

• $< \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} > \mathrm{or} < - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}} >$

• None of the above

A plane meets the axes in A, B and C such that centroid of the $∆$ABC is (1, 2, 3). The equation of the plane is

• x + y/2 + z/3 = 1

• x/3 + y/6 + z/9 = 1

• x + 2y + 3z = 1

• None of these

If $\mathrm{\alpha }, \mathrm{\beta } \mathrm{and} \mathrm{\gamma }$ are the angles which a half ray makes with the positive direction of the axes, then ${\sin }^2\left(\mathrm{\alpha }\right) + {\sin }^2\left(\mathrm{\beta }\right) + {\sin }^2\left(\mathrm{\gamma }\right)$ is equal to

• 1

• 2

• 0

• - 1

The angle between a line with direction ratio 2 : 2 : 1 and a line joining (3, 1, 4) to (7, 2, 12) is

• ${\cos }^{-1}\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

• ${\cos }^{-1}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

• ${\tan }^{-1}\left(\raisebox{1ex}{$- 2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

• None of the above

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

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{6}$

Direction cosines of the line $\frac{\mathrm{x} + 2}{2} = \frac{2\mathrm{y} - 5}{3}, \mathrm{z} = - 1$ are

• $\frac{4}{5}, \frac{3}{5}, 0$

• $\frac{3}{5}, \frac{4}{5}, \frac{1}{5}$

• $- \frac{3}{5}, \frac{4}{5}, 0$

• $\frac{4}{5}, - \frac{2}{5}, \frac{1}{5}$

The acute angle between the line r = $\left(\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + \hat{\mathrm{k}}\right) + \mathrm{\lambda }\left(\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}\right)$ and the plane $\left(2\hat{\mathrm{i}} - \hat{\mathrm{j}} + \hat{\mathrm{k}}\right)$ = 5

• ${\cos }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

• ${\sin }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

• ${\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

• ${\sin }^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$

If A and B are foot of perpendicular drawn from point Q(a, b, c) to the planes yz and zx, then equation of plane through the points A, B and O is

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

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

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

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

If line joining points A and B having position vectors 6a - 4b + 4c and - 4c respectively and the line joining the points C and 0 having position vectors - a - 2b - 3c and a + 2b - 5c intersect, then point of intersection is

• B

• C

• D

• A