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Find the Image Of: (–4, 0, 0) in the Xy-plane. - Mathematics

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प्रश्न

Find the image  of: 

 (–4, 0, 0) in the xy-plane. 

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उत्तर

(-4,0,0)

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अध्याय 28: Introduction to three dimensional coordinate geometry - Exercise 15.1 [पृष्ठ ६]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 28 Introduction to three dimensional coordinate geometry
Exercise 15.1 | Q 2.5 | पृष्ठ ६

वीडियो ट्यूटोरियलVIEW ALL [1]

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).


Coordinate planes divide the space into ______ octants.


A cube of side 5 has one vertex at the point (1, 0, –1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.


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. 


Find the distances of the point P(–4, 3, 5) from the coordinate axes. 


Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).


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


Find the points on z-axis which are at a distance \[\sqrt{21}\]from the point (1, 2, 3). 


If A(–2, 2, 3) and B(13, –3, 13) are two points.
Find the locus of a point P which moves in such a way the 3PA = 2PB.


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. 


Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?


Show that the plane ax + by cz + d = 0 divides the line joining the points (x1y1z1) and (x2y2z2) in the ratio \[- \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}\]


Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.


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).


The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is


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(3, 3,4) from the x-axis is 


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


A plane meets the co-ordinates axis in A, B, C such that the centroid of the ∆ABC is the point (α, β, γ). Show that the equation of the plane is `x/alpha + y/beta + z/γ` = 3


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


Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.


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 equations of the two lines through the origin which intersect the line `(x - 3)/2 = (y - 3)/1 = z/1` at angles of `pi/3` each.


Find the foot of perpendicular from the point (2,3,–8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.


The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is ax + by `+- (sqrt(a^2 + b^2) tan alpha)z ` = 0


The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α 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 angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.


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


If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vecr.(5hati - 3hatj - 2hatk)` = 38.


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