Eighth
To determine the octant in which the point (2, -4, -7) lies, we need to consider the signs of its coordinates.
In an octant, the x, y, and z coordinates can have either positive (+) or negative (-) values.
Given the point (2, -4, -7), we can determine the octant as follows:
First, consider the x-coordinate, which is 2. Since it is positive, we eliminate the octants with negative x-values (4th, 5th, 6th, and 7th).
Next, consider the y-coordinate, which is -4. Since it is negative, we eliminate the octants with positive y-values (1st, 2nd, 5th, and 6th).
Lastly, consider the z-coordinate, which is -7. Since it is negative, we eliminate the octants with positive z-values (1st, 2nd, 3rd, and 4th).
Based on the elimination process, we find that the point (2, -4, -7) lies in the Eighth octant.
Therefore, the octant in which the point (2, -4, -7) lies is the Eighth octant (option A).
The point is (2, -4, -7)
→ x > 0, y < 0, z < 0
In 3D geometry, space is divided into **8 octants** based on the signs of x, y, and z:
So, the point (2, -4, -7) lies in the Eighth octant.
Correct answer: Eighth
The octants in 3D space are defined by the signs of the x, y, and z coordinates.
The signs for each octant are as follows:
The point (2, -4, -7) has signs (+, -, -). Comparing this to the table, we see that it corresponds to Octant VIII (Eighth Octant).
Therefore, the octant in which the point (2, -4, -7) lies is the Eighth octant.
Answer: Eighth
The foot of perpendicular from the origin $O$ to a plane $P$ which meets the co-ordinate axes at the points $A , B , C$ is $(2, a , 4), a \in N$ If the volume of the tetrahedron $OABC$ is 144 unit $^3$, then which of the following points is NOT on $P$ ?
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly: