11th Mathematics NCERT Chapter 11
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– Coordinate Axes and Coordinate Planes in Three Dimensional Space:
In three-dimensional space, the coordinate axes are three mutually perpendicular lines (x, y, and z-axes) that intersect at the origin. The space is divided into eight parts called octants【4:3†source】.
– Points and Octants:
Points in three-dimensional geometry are represented by triplets like (x, y, z), where x, y, and z are the distances from the YZ, ZX, and XY-planes respectively. Different points lie in different octants based on the signs of their coordinates in each octant【4:1†source】.
– Distance between Two Points:
The distance between two points P(x1, y1, z1) and Q(x2, y2, z2) in three-dimensional space is calculated using the formula √((x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2)【4:2†source】.
Students should focus on these key topics from the chapter in the provided file for studying and understanding three-dimensional geometry concepts.
What are the coordinates of the centroid of a triangle with vertices A(3, -5, 7) and B(-1, 7, -6) if the centroid is at the point (1, 1, 1)?
The coordinates of the centroid C are (1, 1, 2).
Are the points (0, 7, 10), (-1, 6, 6), and (-4, 9, 6) the vertices of a right-angled triangle?
Yes, the points (0, 7, 10), (-1, 6, 6), and (-4, 9, 6) are the vertices of a right-angled triangle.
Find the equation of the set of points equidistant from (1, 2, 3) and (3, 2, -1).
The equation is 2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z = 2k^2 – 109.
Determine if the points (-2, 3, 5), (1, 2, 3), and (7, 0, -1) are collinear.
Yes, the points (-2, 3, 5), (1, 2, 3), and (7, 0, -1) are collinear.
Find the equation of the set of points P, such that the sum of their distances from A(4, 0, 0) and B(-4, 0, 0) is equal to 10.
The equation is 2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z = 2k^2 – 109.
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