Advertisements
Advertisements
प्रश्न
The perpendicular distance of the point P (6, 7, 8) from xy - plane is
पर्याय
8
7
6
10
Advertisements
उत्तर
The distance of the point P (6, 7, 8) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 8.
Hence, the correct answer is option (a).
APPEARS IN
संबंधित प्रश्न
Name the octants in which the following points lie: (5, 2, 3)
Name the octants in which the following points lie:
(–5, 4, 3)
Find the image of:
(–2, 3, 4) in the yz-plane.
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 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).
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.
Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.
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.
Verify the following:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
Verify the following:
(5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Find the equation of the set of the points P such that its distances from the points A(3, 4, –5) and B(–2, 1, 4) are equal.
Find the ratio in which the sphere x2 + y2 + z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).
Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).
The length of the perpendicular drawn from the point P(a, b, c) from z-axis is
If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line.
Find the direction cosines of the line passing through the points P(2, 3, 5) and Q(–1, 2, 4).
Find the coordinates of the point where the line through (3, – 4, – 5) and (2, –3, 1) crosses the plane passing through three points (2, 2, 1), (3, 0, 1) and (4, –1, 0)
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).
Find the image of the point having position vector `hati + 3hatj + 4hatk` in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0.
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 angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0
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
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 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.
Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
If the directions cosines of a line are k, k, k, then ______.
The direction cosines of the vector `(2hati + 2hatj - hatk)` are ______.
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 angle between the line `vecr = (5hati - hatj - 4hatk) + lambda(2hati - hatj + hatk)` and the plane `vec.(3hati - 4hatj - hatk)` + 5 = 0 is `sin^-1(5/(2sqrt(91)))`.
The angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
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.
