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Coordinate planes divide the space into ______ octants. - Mathematics

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प्रश्न

Coordinate planes divide the space into ______ octants.

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उत्तर

Coordinate planes divide the space into eight octants.

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अध्याय 12: Introduction to Three Dimensional Geometry - Exercise 12.1 [पृष्ठ २७१]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 12 Introduction to Three Dimensional Geometry
Exercise 12.1 | Q 4.3 | पृष्ठ २७१

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संबंधित प्रश्न

Name the octants in which the following points lie:

(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5),

(–3, –1, 6), (2, –4, –7).


If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.


Name the octants in which the following points lie:

(–5, 4, 3) 


Name the octants in which the following points lie: 

(–5, –4, 7) 


Planes are drawn parallel to the coordinate planes through the points (3, 0, –1) and (–2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.


Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed. 


Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1). 


Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).


Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).


Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle. 


Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.


Find the ratio in which the sphere x2 + y2 z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).


What is the locus of a point for which y = 0, z = 0?


Find the point on y-axis which is at a distance of  \[\sqrt{10}\] units from the point (1, 2, 3).


Let (3, 4, –1) and (–1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to 


What is the locus of a point for which y = 0, z = 0?


The perpendicular distance of the point P (6, 7, 8) from xy - plane is


The length of the perpendicular drawn from the point P(a, b, c) from z-axis is 


Find the direction cosines of the line passing through the points P(2, 3, 5) and Q(–1, 2, 4).


A line makes equal angles with co-ordinate axis. Direction cosines of this line are ______.


If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines are ______.


If a line makes angles α, β, γ with the positive directions of the coordinate axes, then the value of sin2α + sin2β + sin2γ is ______.


Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.


If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane


Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0


O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.


Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.


Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.


The area of the quadrilateral ABCD, where A(0, 4, 1), B(2,  3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to ______.


The direction cosines of the vector `(2hati + 2hatj - hatk)` are ______.


The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.


The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hati + 2/sqrt(14)hatj + 3/sqrt(14)hatk`.


The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.


The line `vecr = 2hati - 3hatj - hatk + lambda(hati - hatj + 2hatk)` lies in the plane `vecr.(3hati + hatj - hatk) + 2` = 0.


The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vecr = (5hati - 4hatj + 6hatk) + lambda(3hati + 7hatj - 2hatk)`.


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