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प्रश्न
Find the image of:
(–5, 4, –3) in the xz-plane.
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उत्तर
(-5,-4,-3)
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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:
(7, 4, –3)
Name the octants in which the following points lie:
(–5, –4, 7)
Name the octants in which the following points lie:
(2, –5, –7)
Find the image of:
(5, 2, –7) in the xy-plane.
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.
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 points on z-axis which are at a distance \[\sqrt{21}\]from the point (1, 2, 3).
Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.
Prove that the point A(1, 3, 0), B(–5, 5, 2), C(–9, –1, 2) and D(–3, –3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.
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.
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
Verify the following:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
Find the ratio in which the sphere x2 + y2 + z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).
Write the distance of the point P (2, 3,5) from the xy-plane.
Find the ratio in which the line segment joining the points (2, 4,5) and (3, −5, 4) is divided by the yz-plane.
Find the point on y-axis which is at a distance of \[\sqrt{10}\] units from the point (1, 2, 3).
The perpendicular distance of the point P(3, 3,4) from the x-axis is
The x-coordinate of a point on the line joining the points Q(2, 2, 1) and R(5, 1, –2) is 4. Find its z-coordinate.
Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).
The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.
A line makes equal angles with co-ordinate axis. Direction cosines of this line are ______.
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.
If a variable line in two adjacent positions has direction cosines l, m, n and l + δl, m + δm, n + δn, show that the small angle δθ between the two positions is given by δθ2 = δl2 + δm2 + δn2
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
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 locus represented by xy + yz = 0 is ______.
The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.
The angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
The line `vecr = 2hati - 3hatj - hatk + lambda(hati - hatj + 2hatk)` lies in the plane `vecr.(3hati + hatj - hatk) + 2` = 0.
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.
